Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]
Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Prove that: \[\cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right) = \sin 2x\]
If \[\cos x = - \frac{3}{5}\] and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .
If 0 ≤ x ≤ π and x lies in the IInd quadrant such that \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan\frac{x}{2}\]
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]
If \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
Prove that \[\left| \cos x \cos \left( \frac{\pi}{3} - x \right) \cos \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.
If \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
The value of `cos^2 48^@ - sin^2 12^@` is ______.
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
