Advertisements
Advertisements
प्रश्न
Find the value of cos 2A, A lies in the first quadrant, when cos A = `15/17`
Advertisements
उत्तर
we know sin2A + cos2A = 1
sin2A = 1 – cos2A
= `1 - (15/17)^2`
= `1 - 225/289`
= `(289 - 225)/289`
sin2A = `64/289`
sin A = `+- sqrt(64/289)`
= `+- 8/17`
Since A lies in the first quadrant, sin A is positive
∴ sin A = `8/17`
cos 2A = cos2A – sin2A
= `(15/17)^2 - 64/289`
=`225/289 - 64/289`
= `(225- 64)/289`
= `161/289`
APPEARS IN
संबंधित प्रश्न
Find the values of sin(480°)
Find the values of cos(300°)
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
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 x = `15/17` and cos y = `12/13, 0 < x < pi/2, 0 < y < pi/2`, find the value of cos(x − y)
If a cos(x + y) = b cos(x − y), show that (a + b) tan x = (a − b) cot y
Show that cos2 A + cos2 B – 2 cos A cos B cos(A + B) = sin2(A + B)
Show that `cot(7 1^circ/2) = sqrt(2) + sqrt(3) + sqrt(4) + sqrt(6)`
Express the following as a product
cos 35° – cos 75°
Show that sin 12° sin 48° sin 54° = `1/8`
Show that `(sin 8x cos x - sin 6x cos 3x)/(cos 2x cos x - sin 3x sin 4x)` = tan 2x
Show that cot(A + 15°) – tan(A – 15°) = `(4cos2"A")/(1 + 2 sin2"A")`
If A + B + C = 180°, prove that sin2A + sin2B + sin2C = 2 + 2 cos A cos B cos C
If A + B + C = 180°, prove that sin2A + sin2B − sin2C = 2 sin A sin B cos C
If A + B + C = 180°, prove that `tan "A"/2 tan "B"/2 + tan "B"/2 tan "C"/2 + tan "C"/2 tan "A"/2` = 1
If x + y + z = xyz, then prove that `(2x)/(1 - x^2) + (2y)/(1 - y^2) + (2z)/(1 - z^2) = (2x)/(1 - x^2) (2y)/(1 - y^2) (2z)/(1 - z^2)`
If A + B + C = `pi/2`, prove the following sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Choose the correct alternative:
If cos 28° + sin 28° = k3, then cos 17° is equal to
