Advertisements
Advertisements
Question
Prove that `(sin x + sin 3x + sin 5x + sin 7x)/(cos x + cos x + cos 5x cos 7x)` = tan 4x
Advertisements
Solution
Nr: (sin x + sin 7x) + (sin 3x + sin 5x)
= `[2sin (7x + x)/2 cos (7x - x)/2] + [2sin (5x + 3x)/2 cos (5x - 3x)/2]`
= 2 sin 4x cos 3x + 2 sin 4x cos x
= 2 sin 4x (cos 3x + cos x) .....(1)
Dr. (cos x + cos 7x) + (cos 3x + cos 5x)
= `[2cos (7x + x)/2 cos (7x - x)/2] + [2cos (5x + 3x)/2 cos (5x - 3x)/2]`
= 2 cos 4x cos 3x + 2 cos 4x cosx
= 2 cos 4x (cos 3x + cos x) .....(2)
L.H.S = `((1))/((2))`
= `(2sin 4x(cos 3x + cos x))/(2cos 4x(cos 3x + cos x))`
= tan 4x
= R.H.S
APPEARS IN
RELATED QUESTIONS
Find the values of tan(1050°)
`(5/7, (2sqrt(6))/7)` is a point on the terminal side of an angle θ in standard position. Determine the six trigonometric function values of angle θ
Find the value of the trigonometric functions for the following:
cos θ = `- 1/2`, θ lies in the III quadrant
Find the value of the trigonometric functions for the following:
tan θ = −2, θ lies in the II quadrant
Prove that sin 105° + cos 105° = cos 45°
Prove that sin 75° – sin 15° = cos 105° + cos 15°
Prove that cos(A + B) cos C – cos(B + C) cos A = sin B sin(C – A)
Show that tan(45° + A) = `(1 + tan"A")/(1 - tan"A")`
Show that tan(45° − A) = `(1 - tan "A")/(1 + tan "A")`
Find the value of cos 2A, A lies in the first quadrant, when tan A `16/63`
Prove that `32(sqrt(3)) sin pi/48 cos pi/48 cos pi/24 cos pi/12 cos pi/6` = 3
Express the following as a sum or difference
cos 5θ cos 2θ
Express the following as a product
cos 35° – cos 75°
Prove that cos(30° – A) cos(30° + A) + cos(45° – A) cos(45° + A) = `cos 2"A" + 1/4`
If A + B + C = 180◦, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos2 B + cos2 C = 1
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that sin2 B + sin2 C = 1
Choose the correct alternative:
If `pi < 2theta < (3pi)/2`, then `sqrt(2 + sqrt(2 + 2cos4theta)` equals to
Choose the correct alternative:
Let fk(x) = `1/"k" [sin^"k" x + cos^"k" x]` where x ∈ R and k ≥ 1. Then f4(x) − f6(x) =
