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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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The value of $\sin (45^{\large\circ}+\theta)-\cos(45^{\large\circ}-\theta)$ is

$\begin{array}{1 1}(A)\;2\cos\theta&(B)\;2\sin \theta\\(C)\;1&(D)\;0\end{array} $

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1 Answer

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Toolbox:
  • $\sin(a+b)=\sin a\cos b+\cos a\sin b$
  • $\cos(a-b)=\cos a\cos b+\sin a \sin b$
$\sin (45^{\large\circ}+\theta)-\cos(45^{\large\circ}-\theta)$
$\sin 45^{\large\circ}\cos\theta+\cos 45^{\large\circ}\sin \theta-\cos 45^{\large\circ}\cos\theta-\sin 45\sin \theta$
$\big(\large\frac{1}{\sqrt 2}$$\cos \theta+\large\frac{1}{\sqrt 2}$$\sin \theta\big)-\big(\large\frac{1}{\sqrt 2}$$\cos \theta+\large\frac{1}{\sqrt 2}$$\sin \theta\big)=0$
Hence (D) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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