Advertisements
Advertisements
प्रश्न
If \[\sin \alpha = \frac{4}{5} \text{ and } \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]
Advertisements
उत्तर
Given:
\[\sin \alpha = \frac{4}{5} \]
\[\cos \beta = \frac{5}{13}\] .
Now,
\[\cos\alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left( \frac{4}{5} \right)^2} = \frac{3}{5}\]
\[\text{ And } , \]
\[sin\beta = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \left( \frac{5}{13} \right)^2} = \frac{12}{13}\]
Now,
\[\cos\left( \alpha - \beta \right) = cos\alpha \times cos\beta + sin\alpha \times sin\beta\]
\[ \Rightarrow \cos\left( \alpha - \beta \right) = \frac{3}{5} \times \frac{5}{13} \times \frac{4}{5} \times \frac{12}{13} = \frac{63}{65}\]
\[\text{ Thus } , \]
\[\cos\frac{\alpha - \beta}{2} = \sqrt{\frac{1 + \cos\left( \alpha - \beta \right)}{2}}\]
\[ = \sqrt{\frac{1 + \frac{63}{65}}{2}}\]
\[ = \frac{8}{\sqrt{65}}\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
If \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan \frac{x}{2}\] .
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
If \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
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: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
If \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
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 \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
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
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]
The value of `cos^2 48^@ - sin^2 12^@` is ______.
The value of sin 20° sin 40° sin 60° sin 80° is ______.
If tanθ = `1/2` and tanΦ = `1/3`, then the value of θ + Φ is ______.
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]`
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
