Advertisements
Advertisements
Question
Show that `sin^2 pi/18 + sin^2 pi/9 + sin^2 (7pi)/18 + sin^2 (4pi)/9` = 2
Advertisements
Solution
LHS = sin2 10° + sin2 20° + sin2 70° + sin2 80°
= sin2 10° + sin2 (90° – 10°) + sin2 20° + sin2(90° – 20°)
= sin2 10° + (cos 10°)2 + sin2 20° + (cos 20°)2
= (sin2 10+ cos2 10) + sin2 20° + cos2 20°
= 1 + 1
= 2
= R.H.S
APPEARS IN
RELATED QUESTIONS
Find the values of `sin (-(11pi)/3)`
`(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
Prove that `(cot(180^circ + theta) sin(90^circ - theta) cos(- theta))/(sin(270^circ + theta) tan(- theta) "cosec"(360^circ + theta))` = cos2θ cotθ
If sin A = `3/5` and cos B = `9/41, 0 < "A" < pi/2, 0 < "B" < pi/2`, find the value of cos(A – B)
Expand cos(A + B + C). Hence prove that cos A cos B cos C = sin A sin B cos C + sin B sin C cos A + sin C sin A cos B, if A + B + C = `pi/2`
Prove that sin(30° + θ) + cos(60° + θ) = cos θ
Prove that cos(A + B) cos C – cos(B + C) cos A = sin B sin(C – A)
Prove that sin2(A + B) – sin2(A – B) = sin2A sin2B
Prove that `tan(pi/4 + theta) tan((3pi)/4 + theta)` = – 1
If θ is an acute angle, then find `cos (pi/4 + theta/2)`, when sin θ = `8/9`
Show that sin 12° sin 48° sin 54° = `1/8`
Prove that 1 + cos 2x + cos 4x + cos 6x = 4 cos x cos 2x cos 3x
If A + B + C = 180°, prove that cos A + cos B − cos C = `- 1 + 4cos "A"/2 cos "B"/2 sin "C"/2`
If A + B + C = 180°, prove that sin A + sin B + sin C = `4 cos "A"/2 cos "B"/2 cos "C"/2`
If A + B + C = 180°, prove that sin(B + C − A) + sin(C + A − B) + sin(A + B − C) = 4 sin A sin B sin C
If ∆ABC is a right triangle and if ∠A = `pi/2` then prove that cos B – cos C = `- 1 + 2sqrt(2) cos "B"/2 sin "C"/2`
Choose the correct alternative:
`1/(cos 80^circ) - sqrt(3)/(sin 80^circ)` =
Choose the correct alternative:
If `pi < 2theta < (3pi)/2`, then `sqrt(2 + sqrt(2 + 2cos4theta)` equals to
