Advertisements
Advertisements
प्रश्न
Prove that: \[\sin^2 42° - \cos^2 78 = \frac{\sqrt{5} + 1}{8}\]
Advertisements
उत्तर
\[LHS = \sin^2 42° - \cos^2 78° \]
\[ = \sin^2 \left( 90° - 48° \right) - \cos^2 \left( 90° - 12° \right)\]
\[ = \cos^2 48° - \sin^2 12° \]
\[ = \cos\left( 48° + 12° \right) \cos\left( 48° - 12° \right) \left[ \cos\left( A + B \right) \cos\left( A - B \right) = \cos^2 A - \sin^2 \right]\]
\[ = \cos60° \cos36° \]
\[ = \frac{1}{2} \times \frac{\sqrt{5} + 1}{4} \left( \because \cos36° = \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{\sqrt{5} + 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
If \[\cos x = - \frac{3}{5}\] and x lies in the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .
If \[\text{ tan } x = \frac{b}{a}\] , then find the value of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\] .
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]
If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(i) \[\tan\alpha + \tan\beta = \frac{2b}{a + c}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
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: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
If \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
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
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is
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 \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\cos \left( 36° - A \right) \cos \left( 36° + A \right) + \cos \left( 54° - A \right) \cos \left( 54° + A \right)\] is
The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
The value of `cos^2 48^@ - sin^2 12^@` is ______.
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
