How do you show that sin(x+60) + sin(x+120) =(√3)cosx ?

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How do you show that sin(x+60) + sin(x+120) =(√3)cosx ?

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Answer 1 · 2018-02-28T09:16:33

See the answer below...

Explanation:

$sin (x + 60) + sin (x + 120)$ $sin(x+60)+sin(x+120)$

$= 2 \cdot sin {\frac{(x + 60) + (x + 120)}{2}} \cdot cos {\frac{(x + 60) - (x + 120)}{2}}$ $=2 cdot sin{((x+60)+(x+120))/2} cdot cos{((x+60)-(x+120))/2}$

$= 2 \cdot sin {\frac{2 x + 180}{2}} \cdot cos {\frac{- 60}{2}}$ $=2 cdot sin{(2x+180)/2} cdot cos{(-60)/2}$

$= 2 \cdot sin {90 + x} \cdot cos {30}$ $=2 cdot sin{90+x} cdot cos{30}$

$= 2 \cdot sin {90 - (- x)} \cdot cos 30$ $=2 cdot sin{90-(-x)} cdot cos30$

$= 2 \cdot cos x \cdot \frac{\sqrt{3}}{2}$ $=2 cdot cosx cdot sqrt3/2$

$= \sqrt{3} \cdot cos x$ $=sqrt3 cdot cosx$

Answer 2 · 2018-02-28T09:25:45

see below

Explanation:

We have, $sin (x + 60) + sin (x + 120)$ $sin(x+60) +sin(x+120)$ on L.H.S
According to formula, $sin (a + b) = sin a cos b + cos a sin b$ $sin(a+b)=sinacosb+cosasinb$
Applying it,we get,
$sin x cos (60) + cos x sin (60) + sin x cos (120) + cos x sin (120)$ $sinxcos(60) +cosxsin(60) +sinxcos(120) + cosxsin(120)$
or, $sin x / 2 + cos x \times (\sqrt{3}) / 2 - sin x / 2 + cos x \times (\sqrt{3}) / 2$ $sinx//2 +cosx xx (sqrt3)//2 -sinx//2 +cosx xx(sqrt3)//2$
{ $a s cos 120 i s - 1 / 2 and sin 120 i s (\sqrt{3}) / 2$ $color(blue)(as cos "120 " is -1//2 and sin "120 " is (sqrt3)//2$ }

Thus,the expression comes out to be, $2 \times (\sqrt{3}) / 2 \times cos x$ $2xx (sqrt3)//2 xxcosx$ which is $\sqrt{3} cos x$ $color(red)(sqrt3 cosx)$ , the R.H.S

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How do you show that sin(x+60) + sin(x+120) =(√3)cosx ?

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