Advertisements
Advertisements
प्रश्न
Find the value of cos 2A, A lies in the first quadrant, when sin A = `4/5`
उत्तर
we know sin2A + cos2A = 1
cos2 A = 1 – sin2A
= `1 - (4/5)^2`
= `1 - 16/25`
= `(25 - 16)/25`
= `9/25`
cos A = `+- sqrt(9/25)`
= `+- 3/5`
Since A lies in the first quadrant, cos A is positive
∴ cos A = `3/5`
cos 2A = cos2A – sin2A
= `(3/5)^2 - (4/5)^2`
= `9/25 - 16/25`
= `(9 - 16)/25`
= `(-7)/25`
APPEARS IN
संबंधित प्रश्न
Find the value of the trigonometric functions for the following:
cos θ = `- 1/2`, θ lies in the III quadrant
Show that `sin^2 pi/18 + sin^2 pi/9 + sin^2 (7pi)/18 + sin^2 (4pi)/9` = 2
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)
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2`, find the value of cos(x − y)
If sin x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2`, find the value of tan(x + y)
If sin A = `3/5` and cos B = `9/41 0 < "A" < pi/2, 0 < "B" < pi/2`, find the value of sin(A + B)
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)
Find sin(x – y), given that sin x = `8/17` with 0 < x < `pi/2`, and cos y = `- 24/25`, x < y < `(3pi)/2`
Find the value of cos 105°.
Prove that cos(π + θ) = − cos θ
Prove that sin(A + B) sin(A – B) = sin2A – sin2B
Show that tan(45° − A) = `(1 - tan "A")/(1 + tan "A")`
Find the value of tan(α + β), given that cot α = `1/2`, α ∈ `(pi, (3pi)/2)` and sec β = `- 5/3` β ∈ `(pi/2, pi)`
If θ is an acute angle, then find `cos (pi/4 + theta/2)`, when sin θ = `8/9`
Prove that (1 + tan 1°)(1 + tan 2°)(1 + tan 3°) ..... (1 + tan 44°) is a multiple of 4
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
sin 4x cos 2x
Express the following as a product
cos 65° + cos 15°
Show that `cos pi/15 cos (2pi)/15 cos (3pi)/15 cos (4pi)/15 cos (5pi)/15 cos (6pi)/15 cos (7pi)/15 = 1/128`
Prove that `(sin(4"A" - 2"B") + sin(4"B" - 2"A"))/(cos(4"A" - 2"B") + cos(4"B" - 2"A"))` = tan(A + B)
