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Algebra Difficulty 3.2 AMC 10/12 Find the answer

Given that sin⁡(π+x)+cos⁡(π+x)=12\sin (π+x)+ \cos (π+x)= \frac {1}{2}, find the value of sin⁡2x=\sin 2x= \_\_\_\_\_\_, and the value of 1+tan⁡xsin⁡xcos⁡(x−π4)=\frac {1+\tan x}{\sin x\cos (x- \frac {π}{4})}= \_\_\_\_\_\_.

A number or a short expression. Spacing and $ signs are ignored.

Solution

We have sin⁡(π+x)+cos⁡(π+x)=−sin⁡x−cos⁡x=12\sin (π+x)+ \cos (π+x)= -\sin x- \cos x= \frac {1}{2}, which implies that sin⁡x+cos⁡x=−12\sin x+ \cos x=- \frac {1}{2}.
Squaring both sides, we get sin⁡2x+2sin⁡xcos⁡x+cos⁡2x=14\sin ^{2}x+2\sin x\cos x+ \cos ^{2}x= \frac {1}{4}, which simplifies to 1+sin⁡2x=141+ \sin 2x= \frac {1}{4}.
Thus, sin⁡2x=−34\sin 2x=- \frac {3}{4}.

For the second part, we have
1+tan⁡xsin⁡xcos⁡(x−π4)=1+sin⁡xcos⁡x22sin⁡x(cos⁡x+sin⁡x)=2sin⁡xcos⁡x=22sin⁡2x=22−34=−823\frac {1+\tan x}{\sin x\cos (x- \frac {π}{4})}= \frac {1+ \frac {\sin x}{\cos x}}{ \frac {\sqrt {2}}{2}\sin x(\cos x+\sin x)} = \frac {\sqrt {2}}{\sin x\cos x} = \frac {2\sqrt {2}}{\sin 2x} = \frac {2\sqrt {2}}{-\frac {3}{4}} = -\frac {8\sqrt {2}}{3}.

Hence, the solutions are sin⁡2x=−34\boxed{\sin 2x = -\frac{3}{4}} and 1+tan⁡xsin⁡xcos⁡(x−π4)=−823\boxed{\frac {1+\tan x}{\sin x\cos (x- \frac {π}{4})} = -\frac {8\sqrt {2}}{3}}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.