Advertisements
Advertisements
प्रश्न
If \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .
Advertisements
उत्तर
Given: \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\]
\[\cos^2 \alpha + \cos^2 \beta + 2\cos\alpha \cos\beta = 0 . . . . . \left( 1 \right)\]
Also,
\[\sin\alpha + \sin\beta = 0\]
\[ \Rightarrow \cos^2 \alpha - \sin^2 \alpha + \cos^2 \beta - \sin^2 \beta + 2\left( \cos\alpha \cos\beta - \sin\alpha \sin\beta \right) = 0\]
\[ \Rightarrow \cos2\alpha + \cos2\beta + 2\cos\left( \alpha + \beta \right) = 0\]
\[ \Rightarrow \cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{2 + \sqrt{2 + 2 \cos 4x}} = 2 \text{ cos } x\]
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: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
Prove that: \[\cot^2 x - \tan^2 x = 4 \cot 2 x \text{ cosec } 2 x\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
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: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
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 \[\sin \alpha = \frac{4}{5} \text{ and } \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
If \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] 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 \[\lambda =\]
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
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 ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]
Find the value of the expression `cos^4 pi/8 + cos^4 (3pi)/8 + cos^4 (5pi)/8 + cos^4 (7pi)/8`
[Hint: Simplify the expression to `2(cos^4 pi/8 + cos^4 (3pi)/8) = 2[(cos^2 pi/8 + cos^2 (3pi)/8)^2 - 2cos^2 pi/8 cos^2 (3pi)/8]`
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
The value of `sin pi/10 sin (13pi)/10` is ______.
`["Hint: Use" sin18^circ = (sqrt5 - 1)/4 "and" cos36^circ = (sqrt5 + 1)/4]`
