Advertisements
Advertisements
प्रश्न
If θ lies in the first quadrant and cosθ = `8/17`, then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).
Advertisements
उत्तर
cosθ = `8/17`
sinθ = `+- sqrt(1 – cos^2theta)`
Since θ lies in the first quadrant, only a positive sign can be considered.
⇒ sinθ = `sqrt(1 – 64/289)`
= `15/17`
Let, y = cos(30° + θ) + cos(45° – θ) + cos(120° – θ)
We know that,
cos(x + y) = cosx cosy – sinx siny
Therefore,
y = cos30° cosθ – sin30° sinθ + cos45° cosθ + sin45°sinθ + cos120° cosθ + sin120° sinθ
= `sqrt3/2(costheta + sintheta) - 1/2(costheta + sintheta) + 1/sqrt2(costheta + sintheta)`
= `(sqrt3/2 - 1/2 + 1/sqrt2)(costheta + sintheta)`
= `(sqrt3/2 - 1/2 + 1/sqrt2)(8/17 + 15/17)`
on solving,
= `((sqrt3 - 1)/2 + 1/sqrt2)(23/17)`
= `(23/17)((sqrt3 - 1)/2 + 1/sqrt2)`
= `23/17((sqrt3 - 1)/2 + 1/sqrt2)`
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
Prove that: \[\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} + \cos^2 \frac{5\pi}{8} + \cos^2 \frac{7\pi}{8} = 2\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x \text{ cosec } 2 x\]
Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
If \[\text{ sin } x + \text{ cos } x = a\], then find the value of
The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\] is
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\] and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]
The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is
The value of `cos^2 48^@ - sin^2 12^@` is ______.
The greatest value of sin x cos x is ______.
The value of sin 20° sin 40° sin 60° sin 80° is ______.
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
If tanθ = `1/2` and tanΦ = `1/3`, then the value of θ + Φ is ______.
The value of sin50° – sin70° + sin10° is equal to ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
