Find the value of $sin (120^{\circ} - 45^{\circ})$ $sin(120^@-45^@)$ ?

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Find the value of $sin (120^{\circ} - 45^{\circ})$ $sin(120^@-45^@)$ ?

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Answer 1 · 2017-03-31T05:20:56

$sin (120^{\circ} - 45^{\circ}) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ $sin(120^@-45^@)=(sqrt3+1)/(2sqrt2)$

Explanation:

$sin (A - B) = sin A cos B - cos A sin B$ $sin(A-B)=sinAcosB-cosAsinB$

Here if $A = 120^{\circ}$ $A=120^@$ and $B = 45^{\circ}$ $B=45^@$ , we can find $sin (120^{\circ} - 45^{\circ})$ $sin(120^@-45^@)$ ,

but we need to know sine and cosine ratios of the two angles.

Now as $sin (180^{\circ} - A) = sin A$ $sin(180^@-A)=sinA$ but $cos (180^{\circ} - A) = - cos A$ $cos(180^@-A)=-cosA$

as such ${sin 120}^{\circ} = {sin 60}^{\circ} = \frac{\sqrt{3}}{2}$ $sin120^@=sin60^@=sqrt3/2$ and ${cos 120}^{\circ} = - {cos 60}^{\circ} = - \frac{1}{2}$ $cos120^@=-cos60^@=-1/2$ and as ${sin 45}^{\circ} = {cos 45}^{\circ} = \frac{1}{\sqrt{2}}$ $sin45^@=cos45^@=1/sqrt2$

Hence $sin (120^{\circ} - 45^{\circ})$ $sin(120^@-45^@)$

$= {sin 120}^{\circ} {cos 45}^{\circ} - {cos 120}^{\circ} {sin 45}^{\circ}$ $=sin120^@cos45^@-cos120^@sin45^@$

$= \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} - (- \frac{1}{2}) \times \frac{1}{\sqrt{2}}$ $=sqrt3/2xx1/sqrt2-(-1/2)xx1/sqrt2$

$= \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}$ $=sqrt3/(2sqrt2)+1/(2sqrt2)$

$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ $=(sqrt3+1)/(2sqrt2)$

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Find the value of $sin (120^{\circ} - 45^{\circ})$ $sin(120^@-45^@)$ ?

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