Advertisements
Advertisements
Question
Prove that `32(sqrt(3)) sin pi/48 cos pi/48 cos pi/24 cos pi/12 cos pi/6` = 3
Advertisements
Solution
`32sqrt(3)[sin pi/48 xx cos pi/48] = 16sqrt(3)[2sin pi/48 cos pi/48]`
= `16sqrt(3) sin pi/24((2pi)/48 = pi/24)`
Now `16sqrt(3)[sin pi/24 xx cos pi/24]`
= `8sqrt(3)[2 sin pi/24 cos pi/24]`
= `8sqrt(3)[sin (2pi)/24]`
= `8sqrt(3) sin pi/12`
Now `8sqrt(3)[sin pi/12 cos pi/12]`
= `4sqrt(3)[2 sin pi/12 cos pi/12]`
= `4sqrt(3)[sin (2pi)/12]`
= `4sqrt(3)(sin pi/6)`
Now `4sqrt(3) sin pi/6 cos pi/6 = 2sqrt(3)[2sin pi/6 cos pi/6]`
`2sqrt(3)[sin (2pi)/6] = 2sqrt(3) sin pi/3`
= `2sqrt(3) xx sqrt(3)/2`
= 3
= R.H.S
APPEARS IN
RELATED QUESTIONS
Find the values of sin(480°)
Find the values of `tan ((19pi)/3)`
Find the value of the trigonometric functions for the following:
cos θ = `- 1/2`, θ lies in the III quadrant
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2` find the value of sin(x + y)
Find the value of sin105°.
Prove that cos(π + θ) = − cos θ
Prove that sin(45° + θ) – sin(45° – θ) = `sqrt(2) sin θ`
If a cos(x + y) = b cos(x − y), show that (a + b) tan x = (a − b) cot y
If x cos θ = `y cos (theta + (2pi)/3) = z cos (theta + (4pi)/3)`. find the value of xy + yz + zx
Show that cos2 A + cos2 B – 2 cos A cos B cos(A + B) = sin2(A + B)
Prove that `tan(pi/4 + theta) tan((3pi)/4 + theta)` = – 1
Prove that sin 4α = `4 tan alpha (1 - tan^2alpha)/(1 + tan^2 alpha)^2`
Prove that (1 + sec 2θ)(1 + sec 4θ) ... (1 + sec 2nθ) = tan 2nθ
Show that `((cos theta -cos 3theta)(sin 8theta + sin 2theta))/((sin 5theta - sin theta) (cos 4theta - cos 6theta))` = 1
Prove that `sin theta/2 sin (7theta)/2 + sin (3theta)/2 sin (11theta)/2` = sin 2θ sin 5θ
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 A + B + C = `pi/2`, prove the following sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
If A + B + C = `pi/2`, prove the following cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Choose the correct alternative:
`(sin("A" - "B"))/(cos"A" cos"B") + (sin("B" - "C"))/(cos"B" cos"C") + (sin("C" - "A"))/(cos"C" cos"A")` is
