Trigonometric FunctionsClass 11 Mathematics NCERT Solutions

We know that 180° = π radian. Therefore, to convert degrees to radians, we multiply the degree measure by π/180.

Radian measure = 25 × (π/180) = 5π/36

First, we convert 30' to degrees. Since 60' = 1°, 30' = (30/60)° = 0.5°. So, -47° 30' = -(47 + 0.5)° = -47.5° = -95/2 degrees.

Radian measure = (-95/2) × (π/180) = -19π/72

Radian measure = 240 × (π/180) = 4π/3

Radian measure = 520 × (π/180) = 26π/9

We know that π radian = 180°. Therefore, to convert radians to degrees, we multiply the radian measure by 180/π.

Degree measure = (11/16) × (180/π) = (11/16) × (180 × 7 / 22) = (1/16) × (180 × 7 / 2) = (90 × 7) / 16 = (45 × 7) / 8 = 315/8 degrees. 315/8° = 39 3/8° = 39° + (3/8) × 60' = 39° + 45/2' = 39° + 22 1/2' = 39° + 22' + (1/2) × 60'' = 39° 22' 30''.

Degree measure = 39° 22' 30''

Degree measure = -4 × (180/π) = -4 × (180 × 7 / 22) = -4 × (90 × 7 / 11) = -2520/11 degrees. -2520/11° = -229 1/11° = - (229° + (1/11) × 60') = - (229° + 5 5/11') = - (229° + 5' + (5/11) × 60'') = - (229° 5' 27.27'') ≈ -229° 5' 27'' (approximately).

Degree measure = -229° 5' 27'' (approximately)

Degree measure = (5π/3) × (180/π) = 5 × 60 = 300°

Degree measure = (7π/6) × (180/π) = 7 × 30 = 210°

The number of revolutions made by the wheel in 1 minute (60 seconds) is 360.

Number of revolutions in 1 second = 360 / 60 = 6 revolutions.

We know that in one complete revolution, the angle turned is 2π radians.

Therefore, the angle turned in 6 revolutions = 6 × 2π radians.

Angle turned in 1 second = 12π radians.

Given: Radius of the circle, r = 100 cm Length of the arc, l = 22 cm

We know the relationship between the angle (θ in radians), arc length (l), and radius (r) is θ = l/r.

θ = 22 / 100 radians.

To convert this angle to degrees, we multiply by 180/π.

Degree measure = (22/100) × (180/π) Using π = 22/7: Degree measure = (22/100) × (180 × 7 / 22) = (1/100) × (180 × 7) = 1260 / 100 = 12.6°

Now, we convert the decimal part of the degree into minutes. 0.6° = 0.6 × 60' = 36'

So, the degree measure of the angle is 12° 36'.

Given: Diameter of the circle = 40 cm Radius of the circle, r = Diameter / 2 = 40 / 2 = 20 cm. Length of the chord = 20 cm.

Consider the triangle formed by the chord and the two radii connecting the endpoints of the chord to the center of the circle. Let the center be O and the endpoints of the chord be A and B. In ΔOAB: OA = radius = 20 cm OB = radius = 20 cm AB = length of the chord = 20 cm

Since all three sides of ΔOAB are equal, it is an equilateral triangle. The angle subtended by the chord at the center, θ = 60°.

To find the length of the minor arc, we use the formula l = rθ, where θ must be in radians.

First, convert the angle to radians: θ = 60° = 60 × (π/180) = π/3 radians.

Now, calculate the arc length: l = rθ = 20 × (π/3) = 20π/3 cm.

Let the two circles have radii r₁ and r₂, respectively. Let the length of the arc be 'l' for both circles.

For the first circle: Angle, θ₁ = 60° Radius = r₁

For the second circle: Angle, θ₂ = 75° Radius = r₂

We know the formula l = rθ, where θ must be in radians. First, convert the angles to radians: θ₁ = 60° = 60 × (π/180) = π/3 radians. θ₂ = 75° = 75 × (π/180) = 5π/12 radians.

For the first circle, l = r₁θ₁ = r₁(π/3). For the second circle, l = r₂θ₂ = r₂(5π/12).

Since the arc lengths are the same: r₁(π/3) = r₂(5π/12)

To find the ratio r₁ : r₂, we rearrange the equation: r₁/r₂ = (5π/12) / (π/3) r₁/r₂ = (5π/12) × (3/π) r₁/r₂ = 15/12 = 5/4

Therefore, the ratio of their radii, r₁ : r₂ is 5 : 4.

The length of the pendulum acts as the radius (r) of the circle, and the distance the tip moves is the arc length (l). Given, r = 75 cm.

The angle in radians (θ) is given by the formula θ = l/r.

θ = 10 / 75 = 2/15 radians

θ = 15 / 75 = 1/5 radians

θ = 21 / 75 = 7/25 radians

Given: cos x = -1/2 and x lies in the third quadrant. In the third quadrant, tan x and cot x are positive, while sin x, cos x, sec x, and cosec x are negative.

Given: sin x = 3/5 and x lies in the second quadrant. In the second quadrant, sin x and cosec x are positive, while cos x, tan x, sec x, and cot x are negative.

Given: cot x = 3/4 and x lies in the third quadrant. In the third quadrant, tan x and cot x are positive, while sin x, cos x, sec x, and cosec x are negative.

Given: sec x = 13/5 and x lies in the fourth quadrant. In the fourth quadrant, cos x and sec x are positive, while sin x, tan x, cosec x, and cot x are negative.

Given: tan x = -5/12 and x lies in the second quadrant. In the second quadrant, sin x and cosec x are positive, while cos x, tan x, sec x, and cot x are negative.

We know that the values of sin x repeat after an interval of 360° or 2π radians. We can write 765° as a multiple of 360° plus a remainder.

765° = 2 × 360° + 45° = 720° + 45°

Therefore, sin(765°) = sin(2 × 360° + 45°) = sin(45°).

Since sin(45°) = 1/√2, sin(765°) = 1/√2.

We know that cosec(-x) = -cosec(x). Also, the values of cosec x repeat after an interval of 360°.

cosec(-1410°) = -cosec(1410°)

Now, we write 1410° as a multiple of 360° plus a remainder. 1410° = 3 × 360° + 330° = 1080° + 330° Alternatively, 1410° = 4 × 360° - 30° = 1440° - 30°

Using the second form: -cosec(1410°) = -cosec(4 × 360° - 30°) = -cosec(-30°)

Since cosec(-x) = -cosec(x): -cosec(-30°) = -(-cosec(30°)) = cosec(30°).

We know that cosec(30°) = 2.

Therefore, cosec(-1410°) = 2.

We know that the values of tan x repeat after an interval of π radians. We can write 19π/3 as a multiple of π plus a remainder.

19π/3 = (18π + π)/3 = 18π/3 + π/3 = 6π + π/3

Therefore, tan(19π/3) = tan(6π + π/3). Since tan(nπ + x) = tan(x) for any integer n: tan(6π + π/3) = tan(π/3).

We know that tan(π/3) = √3.

Therefore, tan(19π/3) = √3.

We know that sin(-x) = -sin(x). Also, the values of sin x repeat after an interval of 2π radians.

sin(-11π/3) = -sin(11π/3)

Now, we write 11π/3 as a multiple of 2π. 11π/3 = (12π - π)/3 = 12π/3 - π/3 = 4π - π/3 = 2 × 2π - π/3

-sin(11π/3) = -sin(4π - π/3) Since sin(2nπ + x) = sin(x): -sin(4π - π/3) = -sin(-π/3)

Using sin(-x) = -sin(x): -sin(-π/3) = -(-sin(π/3)) = sin(π/3).

We know that sin(π/3) = √3/2.

Therefore, sin(-11π/3) = √3/2.

We know that cot(-x) = -cot(x). Also, the values of cot x repeat after an interval of π radians.

cot(-15π/4) = -cot(15π/4)

Now, we write 15π/4 as a multiple of π. 15π/4 = (16π - π)/4 = 16π/4 - π/4 = 4π - π/4

-cot(15π/4) = -cot(4π - π/4) Since cot(nπ + x) = cot(x): -cot(4π - π/4) = -cot(-π/4)

Using cot(-x) = -cot(x): -cot(-π/4) = -(-cot(π/4)) = cot(π/4).

We know that cot(π/4) = 1.

Therefore, cot(-15π/4) = 1.

To prove the identity, we will evaluate the Left Hand Side (L.H.S.) by substituting the known values of the trigonometric functions.

L.H.S. = sin²(π/6) + cos²(π/3) - tan²(π/4)

We know the values:

Substitute these values into the L.H.S.: L.H.S. = (1/2)² + (1/2)² - (1)² = 1/4 + 1/4 - 1 = 2/4 - 1 = 1/2 - 1 = -1/2

Since L.H.S. = -1/2 and the Right Hand Side (R.H.S.) is -1/2, we have: L.H.S. = R.H.S.

Hence, proved.

To prove the identity, we will evaluate the Left Hand Side (L.H.S.).

L.H.S. = 2sin²(π/6) + cosec²(7π/6)cos²(π/3)

First, let us find the values of the trigonometric functions:

Now substitute these values into the L.H.S.: L.H.S. = 2(1/2)² + (-2)²(1/2)² = 2(1/4) + (4)(1/4) = 1/2 + 1 = 3/2

Since L.H.S. = 3/2 and the Right Hand Side (R.H.S.) is 3/2, we have: L.H.S. = R.H.S.

Hence, proved.

To prove the identity, we will evaluate the Left Hand Side (L.H.S.).

L.H.S. = cot²(π/6) + cosec(5π/6) + 3tan²(π/6)

First, let us find the values of the trigonometric functions:

Now substitute these values into the L.H.S.: L.H.S. = (√3)² + 2 + 3(1/√3)² = 3 + 2 + 3(1/3) = 3 + 2 + 1 = 6

Since L.H.S. = 6 and the Right Hand Side (R.H.S.) is 6, we have: L.H.S. = R.H.S.

Hence, proved.

To prove the identity, we will evaluate the Left Hand Side (L.H.S.).

L.H.S. = 2sin²(3π/4) + 2cos²(π/4) + 2sec²(π/3)

First, let us find the values of the trigonometric functions:

Now substitute these values into the L.H.S.: L.H.S. = 2(1/√2)² + 2(1/√2)² + 2(2)² = 2(1/2) + 2(1/2) + 2(4) = 1 + 1 + 8 = 10

Since L.H.S. = 10 and the Right Hand Side (R.H.S.) is 10, we have: L.H.S. = R.H.S.

Hence, proved.

We can write 75° as the sum of two standard angles, 45° and 30°. sin 75° = sin(45° + 30°)

Using the identity sin(A + B) = sin A cos B + cos A sin B: sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (1/√2) × (√3/2) + (1/√2) × (1/2) = √3/(2√2) + 1/(2√2) = (√3 + 1) / (2√2)

So, sin 75° = (√3 + 1) / (2√2).

We can write 15° as the difference of two standard angles, 45° and 30°. tan 15° = tan(45° - 30°)

Using the identity tan(A - B) = (tan A - tan B) / (1 + tan A tan B): tan(45° - 30°) = (tan 45° - tan 30°) / (1 + tan 45° tan 30°) = (1 - 1/√3) / (1 + 1 × 1/√3) = ((√3 - 1)/√3) / ((√3 + 1)/√3) = (√3 - 1) / (√3 + 1)

To rationalize the denominator, we multiply the numerator and denominator by (√3 - 1): = [(√3 - 1)(√3 - 1)] / [(√3 + 1)(√3 - 1)] = (√3 - 1)² / ( (√3)² - 1²) = (3 - 2√3 + 1) / (3 - 1) = (4 - 2√3) / 2 = 2 - √3

So, tan 15° = 2 - √3.

We will use the trigonometric identity: cos A cos B - sin A sin B = cos(A + B)

Let A = (π/4 - x) and B = (π/4 - y).

The Left Hand Side (L.H.S.) of the given equation is in the form of this identity. L.H.S. = cos(π/4 - x)cos(π/4 - y) - sin(π/4 - x)sin(π/4 - y)

Applying the identity, we get: L.H.S. = cos[(π/4 - x) + (π/4 - y)] = cos[π/4 + π/4 - x - y] = cos[2π/4 - (x + y)] = cos[π/2 - (x + y)]

Now, we use the co-function identity: cos(π/2 - θ) = sin θ. Let θ = (x + y). L.H.S. = sin(x + y)

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) using the tangent sum and difference identities:

L.H.S. = tan(π/4 + x) / tan(π/4 - x)

First, let's evaluate the numerator, tan(π/4 + x): tan(π/4 + x) = (tan(π/4) + tan x) / (1 - tan(π/4)tan x) Since tan(π/4) = 1: = (1 + tan x) / (1 - tan x)

Next, let's evaluate the denominator, tan(π/4 - x): tan(π/4 - x) = (tan(π/4) - tan x) / (1 + tan(π/4)tan x) = (1 - tan x) / (1 + tan x)

Now, substitute these back into the L.H.S.: L.H.S. = [ (1 + tan x) / (1 - tan x) ] / [ (1 - tan x) / (1 + tan x) ] = [ (1 + tan x) / (1 - tan x) ] × [ (1 + tan x) / (1 - tan x) ] = (1 + tan x)² / (1 - tan x)² = ((1 + tan x) / (1 - tan x))²

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

To prove the identity, we will simplify the Left Hand Side (L.H.S.) using the following properties of trigonometric functions:

L.H.S. = (cos(π+x)cos(-x)) / (sin(π-x)cos(π/2+x))

Substitute the identities into the L.H.S.: L.H.S. = ((-cos x)(cos x)) / ((sin x)(-sin x)) = (-cos²x) / (-sin²x) = cos²x / sin²x

Since cot x = cos x / sin x, we have cot²x = cos²x / sin²x. L.H.S. = cot²x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

To prove the identity, we will simplify the Left Hand Side (L.H.S.) using the following properties of trigonometric functions:

L.H.S. = cos(3π/2+x)cos(2π+x)[cot(3π/2-x) + cot(2π+x)]

Substitute the identities into the L.H.S.: L.H.S. = (sin x)(cos x)[tan x + cot x]

Now, express tan x and cot x in terms of sin x and cos x: tan x = sin x / cos x cot x = cos x / sin x

L.H.S. = (sin x cos x) [ (sin x / cos x) + (cos x / sin x) ]

Find a common denominator for the terms inside the bracket: L.H.S. = (sin x cos x) [ (sin²x + cos²x) / (cos x sin x) ]

Using the Pythagorean identity sin²x + cos²x = 1: L.H.S. = (sin x cos x) [ 1 / (cos x sin x) ]

Cancel out the (sin x cos x) terms: L.H.S. = 1

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will use the trigonometric identity: cos A cos B + sin A sin B = cos(A - B)

The Left Hand Side (L.H.S.) of the given equation is in the form of this identity. L.H.S. = sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x

Let A = (n+2)x and B = (n+1)x. Rewriting the L.H.S. to match the identity form: L.H.S. = cos((n+2)x)cos((n+1)x) + sin((n+2)x)sin((n+1)x)

Applying the identity cos(A - B), we get: L.H.S. = cos[ (n+2)x - (n+1)x ] = cos[ (n+2 - (n+1))x ] = cos[ (n + 2 - n - 1)x ] = cos[ (1)x ] = cos x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will use the sum-to-product identity: cos A - cos B = -2 sin((A + B)/2) sin((A - B)/2)

Let A = (3π/4 + x) and B = (3π/4 - x).

The Left Hand Side (L.H.S.) of the given equation is in the form of this identity. L.H.S. = cos(3π/4 + x) - cos(3π/4 - x)

First, we find (A + B)/2 and (A - B)/2:

Now, apply the identity: L.H.S. = -2 sin(3π/4) sin(x)

We need to find the value of sin(3π/4). 3π/4 is in the second quadrant, where sine is positive. sin(3π/4) = sin(π - π/4) = sin(π/4) = 1/√2.

Substitute this value back into the L.H.S.: L.H.S. = -2 (1/√2) sin x = - (2/√2) sin x = -√2 sin x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will use the identity: sin²A - sin²B = sin(A + B)sin(A - B)

Let A = 6x and B = 4x.

The Left Hand Side (L.H.S.) of the given equation is in the form of this identity. L.H.S. = sin²6x - sin²4x

Applying the identity, we get: L.H.S. = sin(6x + 4x) sin(6x - 4x) = sin(10x) sin(2x)

This can be rewritten as sin 2x sin 10x, which is the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will first transform the Left Hand Side (L.H.S.) using the Pythagorean identity cos²θ = 1 - sin²θ.

L.H.S. = cos²2x - cos²6x = (1 - sin²2x) - (1 - sin²6x) = 1 - sin²2x - 1 + sin²6x = sin²6x - sin²2x

Now, we use the identity: sin²A - sin²B = sin(A + B)sin(A - B)

Let A = 6x and B = 2x. L.H.S. = sin(6x + 2x) sin(6x - 2x) = sin(8x) sin(4x)

This can be rewritten as sin 4x sin 8x, which is the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will rearrange the Left Hand Side (L.H.S.) and use the sum-to-product identity: sin A + sin B = 2 sin((A + B)/2) cos((A - B)/2)

L.H.S. = sin 2x + 2sin 4x + sin 6x = (sin 6x + sin 2x) + 2sin 4x

Apply the identity to the term in the parenthesis with A = 6x and B = 2x: L.H.S. = [ 2 sin((6x + 2x)/2) cos((6x - 2x)/2) ] + 2sin 4x = [ 2 sin(8x/2) cos(4x/2) ] + 2sin 4x = 2 sin(4x) cos(2x) + 2sin 4x

Now, factor out the common term 2sin 4x: L.H.S. = 2sin 4x (cos 2x + 1)

Next, use the double-angle identity for cosine: cos 2x = 2cos²x - 1. L.H.S. = 2sin 4x ( (2cos²x - 1) + 1 ) = 2sin 4x (2cos²x) = 4cos²x sin 4x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify both the Left Hand Side (L.H.S.) and the Right Hand Side (R.H.S.) separately using the sum-to-product identities:

Simplifying L.H.S.: L.H.S. = cot 4x (sin 5x + sin 3x) Apply the first identity with A = 5x and B = 3x: L.H.S. = cot 4x [ 2 sin((5x + 3x)/2) cos((5x - 3x)/2) ] = cot 4x [ 2 sin(4x) cos(x) ] Express cot 4x as cos 4x / sin 4x: L.H.S. = (cos 4x / sin 4x) × 2 sin 4x cos x = 2 cos 4x cos x

Simplifying R.H.S.: R.H.S. = cot x (sin 5x - sin 3x) Apply the second identity with A = 5x and B = 3x: R.H.S. = cot x [ 2 cos((5x + 3x)/2) sin((5x - 3x)/2) ] = cot x [ 2 cos(4x) sin(x) ] Express cot x as cos x / sin x: R.H.S. = (cos x / sin x) × 2 cos 4x sin x = 2 cos x cos 4x

Since L.H.S. = 2 cos 4x cos x and R.H.S. = 2 cos 4x cos x, we have: L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) using the sum-to-product identities:

L.H.S. = (cos 9x - cos 5x) / (sin 17x - sin 3x)

Numerator: cos 9x - cos 5x = -2 sin((9x + 5x)/2) sin((9x - 5x)/2) = -2 sin(14x/2) sin(4x/2) = -2 sin(7x) sin(2x)

Denominator: sin 17x - sin 3x = 2 cos((17x + 3x)/2) sin((17x - 3x)/2) = 2 cos(20x/2) sin(14x/2) = 2 cos(10x) sin(7x)

Now, substitute the simplified numerator and denominator back into the L.H.S.: L.H.S. = (-2 sin(7x) sin(2x)) / (2 cos(10x) sin(7x))

Cancel the common terms (2 and sin(7x)): L.H.S. = -sin(2x) / cos(10x)

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) using the sum-to-product identities:

L.H.S. = (sin 5x + sin 3x) / (cos 5x + cos 3x)

Numerator: sin 5x + sin 3x = 2 sin((5x + 3x)/2) cos((5x - 3x)/2) = 2 sin(8x/2) cos(2x/2) = 2 sin(4x) cos(x)

Denominator: cos 5x + cos 3x = 2 cos((5x + 3x)/2) cos((5x - 3x)/2) = 2 cos(8x/2) cos(2x/2) = 2 cos(4x) cos(x)

Now, substitute the simplified numerator and denominator back into the L.H.S.: L.H.S. = (2 sin(4x) cos(x)) / (2 cos(4x) cos(x))

Cancel the common terms (2 and cos(x)): L.H.S. = sin(4x) / cos(4x)

Since tan θ = sin θ / cos θ: L.H.S. = tan 4x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) using the sum-to-product identities:

L.H.S. = (sin x - sin y) / (cos x + cos y)

Substitute the identities into the L.H.S.: L.H.S. = [ 2 cos((x + y)/2) sin((x - y)/2) ] / [ 2 cos((x + y)/2) cos((x - y)/2) ]

Cancel the common terms (2 and cos((x + y)/2)): L.H.S. = sin((x - y)/2) / cos((x - y)/2)

Since tan θ = sin θ / cos θ: L.H.S. = tan((x - y)/2)

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) by first rearranging the terms and then using the sum-to-product identities:

L.H.S. = (sin 3x + sin x) / (cos 3x + cos x)

Numerator: sin 3x + sin x = 2 sin((3x + x)/2) cos((3x - x)/2) = 2 sin(4x/2) cos(2x/2) = 2 sin(2x) cos(x)

Denominator: cos 3x + cos x = 2 cos((3x + x)/2) cos((3x - x)/2) = 2 cos(4x/2) cos(2x/2) = 2 cos(2x) cos(x)

Now, substitute the simplified numerator and denominator back into the L.H.S.: L.H.S. = (2 sin(2x) cos(x)) / (2 cos(2x) cos(x))

Cancel the common terms (2 and cos(x)): L.H.S. = sin(2x) / cos(2x)

Since tan θ = sin θ / cos θ: L.H.S. = tan 2x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the numerator and the denominator of the Left Hand Side (L.H.S.) separately.

Numerator: Using the identity sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2): sin x - sin 3x = 2 cos((x + 3x)/2) sin((x - 3x)/2) = 2 cos(4x/2) sin(-2x/2) = 2 cos(2x) sin(-x) Since sin(-θ) = -sin θ: = -2 cos(2x) sin(x)

Denominator: Using the identity cos 2x = cos²x - sin²x: sin²x - cos²x = -(cos²x - sin²x) = -cos(2x)

Now, substitute the simplified numerator and denominator back into the L.H.S.: L.H.S. = (-2 cos(2x) sin(x)) / (-cos(2x))

Cancel the common terms (-1 and cos(2x)): L.H.S. = 2 sin x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

To prove this identity, we will rearrange the terms in the numerator and denominator of the Left Hand Side (L.H.S.) and apply sum-to-product identities.

L.H.S. = (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x) Rearrange the terms: L.H.S. = [(cos 4x + cos 2x) + cos 3x] / [(sin 4x + sin 2x) + sin 3x]

Now, apply the identities:

Numerator: (cos 4x + cos 2x) + cos 3x = [2 cos((4x + 2x)/2) cos((4x - 2x)/2)] + cos 3x = 2 cos(3x) cos(x) + cos 3x Factor out cos 3x: = cos 3x (2 cos x + 1)

Denominator: (sin 4x + sin 2x) + sin 3x = [2 sin((4x + 2x)/2) cos((4x - 2x)/2)] + sin 3x = 2 sin(3x) cos(x) + sin 3x Factor out sin 3x: = sin 3x (2 cos x + 1)

Substitute back into the L.H.S.: L.H.S. = [cos 3x (2 cos x + 1)] / [sin 3x (2 cos x + 1)]

Cancel the common term (2 cos x + 1): L.H.S. = cos 3x / sin 3x

Since cot θ = cos θ / sin θ: L.H.S. = cot 3x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

To prove this identity, we start with the relationship 3x = 2x + x.

Take the cotangent of both sides: cot(3x) = cot(2x + x)

Now, apply the cotangent sum identity: cot(A + B) = (cot A cot B - 1) / (cot B + cot A)

cot(3x) = (cot 2x cot x - 1) / (cot 2x + cot x)

Multiply both sides by (cot 2x + cot x): cot(3x) (cot 2x + cot x) = cot 2x cot x - 1

Distribute cot(3x) on the left side: cot 3x cot 2x + cot 3x cot x = cot 2x cot x - 1

Rearrange the terms to match the required expression: 1 = cot 2x cot x - cot 3x cot 2x - cot 3x cot x

This is the identity we needed to prove. Hence, proved.

We will prove this identity by expressing tan 4x in terms of tan x using the double-angle identity for tangent, tan 2A = (2 tan A) / (1 - tan²A).

First, let's write tan 4x as tan(2 * 2x). Using the identity with A = 2x: tan 4x = (2 tan 2x) / (1 - tan²2x)

Now, we substitute the expression for tan 2x again using the same identity (with A = x): tan 2x = (2 tan x) / (1 - tan²x)

Substitute this into the equation for tan 4x:

Numerator: 2 tan 2x = 2 [ (2 tan x) / (1 - tan²x) ] = (4 tan x) / (1 - tan²x)

Denominator: 1 - tan²2x = 1 - [ (2 tan x) / (1 - tan²x) ]² = 1 - (4 tan²x) / (1 - tan²x)² Find a common denominator: = [ (1 - tan²x)² - 4 tan²x ] / (1 - tan²x)² Expand (1 - tan²x)²: = [ (1 - 2tan²x + tan⁴x) - 4 tan²x ] / (1 - tan²x)² = (1 - 6tan²x + tan⁴x) / (1 - tan²x)²

Now, combine the numerator and denominator for tan 4x: tan 4x = [ (4 tan x) / (1 - tan²x) ] / [ (1 - 6tan²x + tan⁴x) / (1 - tan²x)² ] = [ (4 tan x) / (1 - tan²x) ] × [ (1 - tan²x)² / (1 - 6tan²x + tan⁴x) ] Cancel one (1 - tan²x) term: = (4 tan x (1 - tan²x)) / (1 - 6tan²x + tan⁴x)

This is the Right Hand Side (R.H.S.). Hence, proved.

We will prove this identity by using the double-angle identity for cosine, cos 2A = 1 - 2sin²A.

First, let's write cos 4x as cos(2 * 2x). Using the identity with A = 2x: cos 4x = 1 - 2sin²(2x)

Now, we use the double-angle identity for sine, sin 2x = 2 sin x cos x. Substitute this into the equation: cos 4x = 1 - 2(2 sin x cos x)² = 1 - 2(4 sin²x cos²x) = 1 - 8sin²x cos²x

This is the Right Hand Side (R.H.S.). Hence, proved.

We will prove this identity by using the triple-angle and double-angle identities for cosine.

First, let's write cos 6x as cos(3 * 2x). Using the triple-angle identity with A = 2x: cos 6x = 4cos³(2x) - 3cos(2x)

Now, substitute the expression for cos 2x into this equation: cos 6x = 4(2cos²x - 1)³ - 3(2cos²x - 1)

We need to expand (2cos²x - 1)³. Using the binomial expansion (a - b)³ = a³ - 3a²b + 3ab² - b³: Let a = 2cos²x and b = 1. (2cos²x - 1)³ = (2cos²x)³ - 3(2cos²x)²(1) + 3(2cos²x)(1)² - (1)³ = 8cos⁶x - 3(4cos⁴x) + 6cos²x - 1 = 8cos⁶x - 12cos⁴x + 6cos²x - 1

Now substitute this expansion back into the equation for cos 6x: cos 6x = 4(8cos⁶x - 12cos⁴x + 6cos²x - 1) - 3(2cos²x - 1)

Distribute the constants: cos 6x = (32cos⁶x - 48cos⁴x + 24cos²x - 4) - (6cos²x - 3) cos 6x = 32cos⁶x - 48cos⁴x + 24cos²x - 4 - 6cos²x + 3

Combine like terms: cos 6x = 32cos⁶x - 48cos⁴x + 18cos²x - 1

This is the Right Hand Side (R.H.S.). Hence, proved.

We will use the product-to-sum identity: 2 cos A cos B = cos(A + B) + cos(A - B)

Let's apply this to the first term of the Left Hand Side (L.H.S.). Let A = 9π/13 and B = π/13. 2cos(9π/13)cos(π/13) = cos(9π/13 + π/13) + cos(9π/13 - π/13) = cos(10π/13) + cos(8π/13)

Now, substitute this back into the L.H.S.: L.H.S. = [cos(10π/13) + cos(8π/13)] + cos(3π/13) + cos(5π/13)

We can express some angles in terms of π. Notice that 10π/13 = π - 3π/13 and 8π/13 = π - 5π/13. Using the identity cos(π - θ) = -cos θ:

Substitute these into the L.H.S. expression: L.H.S. = [-cos(3π/13) - cos(5π/13)] + cos(3π/13) + cos(5π/13) = -cos(3π/13) + cos(3π/13) - cos(5π/13) + cos(5π/13) = 0

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will expand the Left Hand Side (L.H.S.) and then group the terms.

L.H.S. = (sin 3x + sin x)sin x + (cos 3x - cos x)cos x

Distribute sin x and cos x: L.H.S. = sin 3x sin x + sin²x + cos 3x cos x - cos²x

Rearrange the terms to group the product terms and the squared terms: L.H.S. = (cos 3x cos x + sin 3x sin x) + (sin²x - cos²x)

Now, we use two identities:

Applying the first identity to the first group with A = 3x and B = x: cos 3x cos x + sin 3x sin x = cos(3x - x) = cos(2x)

Applying the second identity to the second group: sin²x - cos²x = -(cos²x - sin²x) = -cos(2x)

Substitute these back into the L.H.S.: L.H.S. = cos(2x) - cos(2x) = 0

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) using the sum-to-product identities:

L.H.S. = (cos x + cos y)² + (sin x - sin y)²

Substitute the identities into the L.H.S.: L.H.S. = [ 2 cos((x + y)/2) cos((x - y)/2) ]² + [ 2 cos((x + y)/2) sin((x - y)/2) ]²

Square each term: L.H.S. = 4 cos²((x + y)/2) cos²((x - y)/2) + 4 cos²((x + y)/2) sin²((x - y)/2)

Factor out the common term 4 cos²((x + y)/2): L.H.S. = 4 cos²((x + y)/2) [ cos²((x - y)/2) + sin²((x - y)/2) ]

Using the Pythagorean identity cos²θ + sin²θ = 1, where θ = (x - y)/2: L.H.S. = 4 cos²((x + y)/2) [ 1 ] = 4 cos²((x + y)/2)

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the Left Hand Side (L.H.S.) using the sum-to-product identities:

L.H.S. = (cos x - cos y)² + (sin x - sin y)²

Substitute the identities into the L.H.S.: L.H.S. = [ -2 sin((x + y)/2) sin((x - y)/2) ]² + [ 2 cos((x + y)/2) sin((x - y)/2) ]²

Square each term: L.H.S. = 4 sin²((x + y)/2) sin²((x - y)/2) + 4 cos²((x + y)/2) sin²((x - y)/2)

Factor out the common term 4 sin²((x - y)/2): L.H.S. = 4 sin²((x - y)/2) [ sin²((x + y)/2) + cos²((x + y)/2) ]

Using the Pythagorean identity sin²θ + cos²θ = 1, where θ = (x + y)/2: L.H.S. = 4 sin²((x - y)/2) [ 1 ] = 4 sin²((x - y)/2)

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

To prove this identity, we will rearrange the terms on the Left Hand Side (L.H.S.) and apply the sum-to-product identity: sin A + sin B = 2 sin((A + B)/2) cos((A - B)/2).

L.H.S. = sin x + sin 3x + sin 5x + sin 7x

Group the terms strategically: L.H.S. = (sin 7x + sin x) + (sin 5x + sin 3x)

Apply the identity to the first group (A = 7x, B = x): sin 7x + sin x = 2 sin((7x + x)/2) cos((7x - x)/2) = 2 sin(4x) cos(3x)

Apply the identity to the second group (A = 5x, B = 3x): sin 5x + sin 3x = 2 sin((5x + 3x)/2) cos((5x - 3x)/2) = 2 sin(4x) cos(x)

Substitute these back into the L.H.S.: L.H.S. = 2 sin(4x) cos(3x) + 2 sin(4x) cos(x)

Factor out the common term 2 sin(4x): L.H.S. = 2 sin(4x) [cos(3x) + cos(x)]

Now, apply the sum-to-product identity for cosine: cos A + cos B = 2 cos((A + B)/2) cos((A - B)/2). cos(3x) + cos(x) = 2 cos((3x + x)/2) cos((3x - x)/2) = 2 cos(2x) cos(x)

Substitute this back into the L.H.S. expression: L.H.S. = 2 sin(4x) [2 cos(2x) cos(x)] = 4 sin(4x) cos(2x) cos(x)

Rearranging the terms gives: L.H.S. = 4 cos x cos 2x sin 4x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

We will simplify the numerator and denominator of the Left Hand Side (L.H.S.) using sum-to-product identities.

Numerator: Numerator = (sin 7x + sin 5x) + (sin 9x + sin 3x) = [2 sin((7x+5x)/2) cos((7x-5x)/2)] + [2 sin((9x+3x)/2) cos((9x-3x)/2)] = [2 sin(6x) cos(x)] + [2 sin(6x) cos(3x)] Factor out the common term 2 sin(6x): = 2 sin(6x) (cos x + cos 3x)

Denominator: Denominator = (cos 7x + cos 5x) + (cos 9x + cos 3x) = [2 cos((7x+5x)/2) cos((7x-5x)/2)] + [2 cos((9x+3x)/2) cos((9x-3x)/2)] = [2 cos(6x) cos(x)] + [2 cos(6x) cos(3x)] Factor out the common term 2 cos(6x): = 2 cos(6x) (cos x + cos 3x)

Now, substitute the simplified numerator and denominator back into the L.H.S.: L.H.S. = [2 sin(6x) (cos x + cos 3x)] / [2 cos(6x) (cos x + cos 3x)]

Cancel the common terms (2 and (cos x + cos 3x)): L.H.S. = sin(6x) / cos(6x)

Since tan θ = sin θ / cos θ: L.H.S. = tan 6x

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

To prove this identity, we will rearrange the terms on the Left Hand Side (L.H.S.) and apply sum-to-product and double-angle identities.

L.H.S. = sin 3x + sin 2x - sin x

Group the terms: L.H.S. = (sin 3x - sin x) + sin 2x

Apply the identity sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2) to the first group: sin 3x - sin x = 2 cos((3x + x)/2) sin((3x - x)/2) = 2 cos(2x) sin(x)

Substitute this back into the L.H.S.: L.H.S. = 2 cos(2x) sin(x) + sin 2x

Apply the double-angle identity sin 2x = 2 sin x cos x: L.H.S. = 2 cos(2x) sin(x) + 2 sin x cos x

Factor out the common term 2 sin x: L.H.S. = 2 sin x (cos 2x + cos x)

Now, apply the sum-to-product identity cos A + cos B = 2 cos((A + B)/2) cos((A - B)/2): cos 2x + cos x = 2 cos((2x + x)/2) cos((2x - x)/2) = 2 cos(3x/2) cos(x/2)

Substitute this back into the L.H.S. expression: L.H.S. = 2 sin x [2 cos(3x/2) cos(x/2)] = 4 sin x cos(x/2) cos(3x/2)

This is equal to the Right Hand Side (R.H.S.). L.H.S. = R.H.S.

Hence, proved.

Given: tan x = -4/3 and x is in quadrant II. This means π/2 < x < π. Dividing by 2, we get π/4 < x/2 < π/2. This means x/2 is in quadrant I. In quadrant I, sin(x/2), cos(x/2), and tan(x/2) are all positive.

First, find cos x. We know sec²x = 1 + tan²x. sec²x = 1 + (-4/3)² = 1 + 16/9 = 25/9. sec x = ±√(25/9) = ±5/3. Since x is in quadrant II, sec x is negative. So, sec x = -5/3. cos x = 1 / sec x = -3/5.

Now we use the half-angle identities:

Given: cos x = -1/3 and x is in quadrant III. This means π < x < 3π/2. Dividing by 2, we get π/2 < x/2 < 3π/4. This means x/2 is in quadrant II. In quadrant II, sin(x/2) is positive, while cos(x/2) and tan(x/2) are negative.

Now we use the half-angle identities:

Given: sin x = 1/4 and x is in quadrant II. This means π/2 < x < π. Dividing by 2, we get π/4 < x/2 < π/2. This means x/2 is in quadrant I. In quadrant I, sin(x/2), cos(x/2), and tan(x/2) are all positive.

First, find cos x. We know cos²x = 1 - sin²x. cos²x = 1 - (1/4)² = 1 - 1/16 = 15/16. cos x = ±√(15/16) = ±√15/4. Since x is in quadrant II, cos x is negative. So, cos x = -√15/4.

Now we use the half-angle identities: