Let $A = \sin 12^{\circ} + \cos 18^{\circ}$ and $B = \sin 18^{\circ} + \cos 12^{\circ}$. Evaluate $A^2 + B^2 =?$
If $A = \sin 12^{\circ} + \cos 18^{\circ}$ and $B = \sin 18^{\circ} + \cos 12^{\circ}$, $A^2 + B^2 =?$
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trigonometry
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0Any thoughts? Have you tried squaring and adding? – 2017-01-10
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0Well i did everything but the answer must be a concise and easy one instead of a complex solution. So i need a fast and easy way to do this. And the answer is 3. – 2017-01-10
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0Edit your post to include what you get when you square and add. Hint: the addition formulas for the trig functions come in handy. – 2017-01-10
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1$A^2 +B^2 = \sin^212 +\cos^212 +\sin^218 +\cos^218 +2(\sin 12\cos 18 +\sin 18\cos 12) =1+1+2\sin 30 =3$. – 2017-01-10
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0thank you your answer is perfect – 2017-01-10
1 Answers
1
\begin{align*} A &= \sin 12^{\circ}+\cos 18^{\circ} \\ B &= \sin 18^{\circ}+\cos 12^{\circ} \\ A^2 &= \sin^2 12^{\circ}+\cos^2 18^{\circ}+2\sin 12^{\circ} \cos 18^{\circ} \\ B^2 &= \sin^2 18^{\circ}+\cos^2 12^{\circ}+2\sin 18^{\circ} \cos 12^{\circ} \\ A^2+B^2 &= \sin^2 12^{\circ}+\cos^2 12^{\circ}+\sin^2 18^{\circ}+\cos^2 18^{\circ} \\ &\quad +2(\sin 12^{\circ} \cos 18^{\circ}+\sin 18^{\circ} \cos 12^{\circ}) \\ &= 1+1+2\sin 30^{\circ} \\ &= 3 \end{align*}
Useful formulae:
- $\sin^2 \theta+\cos^2 \theta=1$
- $\sin \theta \cos \phi+\cos \theta \sin \phi=\sin (\theta+\phi)$