Math Doubts

Simplify $\sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$

$\sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ is a trigonometric expression and its value can be evaluated by simplifying it. The both trigonometric functions have angles which are sum of two angles so that the value of the trigonometric expression can be evaluated by using angle sum identities.

Expand sine of sum of angles

The trigonometric term $\sin{(\theta+60^°)}$ can be expanded by using sin of sum of two angles trigonometric identity.

$\sin{(\theta+60^°)}$ $\,=\,$ $\sin{\theta}\cos{(60^°)}$ $+$ $\sin{\theta}\cos{(60^°)}$

$\implies \sin{(\theta+60^°)}$ $\,=\,$ $\sin{\theta} \times \cos{(60^°)}$ $+$ $\cos{\theta} \times \sin{(60^°)}$

$\implies \sin{(\theta+60^°)}$ $\,=\,$ $\sin{\theta} \times \dfrac{1}{2}$ $+$ $\cos{\theta} \times \dfrac{\sqrt{3}}{2}$

$\,\,\, \therefore \,\,\,\,\,\, \sin{(\theta+60^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$

Expand cosine of sum of angles

Similarly, the trigonometric term $\cos{(\theta+30^°)}$ can also be expanded by using cos of sum of two angles trigonometric identity.

$\cos{(\theta+30^°)}$ $\,=\,$ $\cos{\theta}\cos{(30^°)}$ $-$ $\sin{\theta}\sin{(30^°)}$

$\implies \cos{(\theta+30^°)}$ $\,=\,$ $\cos{\theta} \times \cos{(30^°)}$ $-$ $\sin{\theta} \times \sin{(30^°)}$

$\implies \cos{(\theta+30^°)}$ $\,=\,$ $\cos{\theta} \times \dfrac{\sqrt{3}}{2}$ $-$ $\sin{\theta} \times \dfrac{1}{2}$

$\,\,\, \therefore \,\,\,\,\,\, \cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$ $-$ $\dfrac{\sin{\theta}}{2}$

Evaluate Trigonometric Expression

Finally, subtract $\cos{(\theta+30^°)}$ from $\sin{(\theta+60^°)}$ to get the value of the trigonometric expression mathematically.

$\sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\Bigg(\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sqrt{3}\cos{\theta}}{2}\Bigg)$ $-$ $\Bigg(\dfrac{\sqrt{3}\cos{\theta}}{2}$ $-$ $\dfrac{\sin{\theta}}{2}\Bigg)$

$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$ $-$ $\dfrac{\sqrt{3}\cos{\theta}}{2}$ $+$ $\dfrac{\sin{\theta}}{2}$

$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\require{\cancel} \cancel{\dfrac{\sqrt{3}\cos{\theta}}{2}}$ $-$ $\require{\cancel} \cancel{\dfrac{\sqrt{3}\cos{\theta}}{2}}$ $+$ $\dfrac{\sin{\theta}}{2}$

$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{\sin{\theta}}{2}$ $+$ $\dfrac{\sin{\theta}}{2}$

$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\dfrac{2\sin{\theta}}{2}$

$\implies \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{2} \sin{\theta}}{\cancel{2}}$

$\,\,\, \therefore \,\,\,\,\,\, \sin{(\theta+60^°)}$ $-$ $\cos{(\theta+30^°)}$ $\,=\,$ $\sin{\theta}$



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