Question

Please help with math questions

Answer 1

Answer:

See answers below

Step-by-step explanation:

Problem 1

Recall that $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$ and that $tan(\frac{\pi}{4})=1$. Using these two facts, we can rewrite the expression:

$\frac{1+tanx}{-1+tanx}\\\\-\frac{1+tanx}{1-tanx}\\ \\-\frac{tan(\frac{\pi}{4})+tanx}{1-tan(\frac{\pi}{4})tan(x)}\\ \\-tan(x+\frac{\pi}{4})\\ \\tan(\frac{3\pi}{4}-x)$

Hence, the first choice is correct

Problem 2

$cos(\frac{x}{2})=\sqrt{3}-cos(\frac{x}{2})\:;\: 0\leq x < 360^\circ\\\\2cos(\frac{x}{2})=\sqrt{3}\\ \\cos(\frac{x}{2})=\frac{\sqrt{3}}{2}\\\\\frac{x}{2}=\frac{\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\ \\ x=\frac{\pi}{3}+4\pi n,\frac{11\pi}{3}+4\pi n\\ \\ x=60^\circ+720n^\circ, 660^\circ+720n^\circ\\\\x=60^\circ$

It's helpful to use the unit circle to solve these kinds of problems. Therefore, the third answer is correct.

Problem 3

Because $sin\theta=-\frac{5}{13}$ and our parameters are $\pi < \theta < \frac{3\pi}{2}$, the triangle must be in Quadrant III where $sin\theta < 0$ and $cos\theta < 0$.

You may recall the double angle formula $sin2\theta=2sin\theta cos\theta$. We can find $cos\theta$ using $sin\theta$ with the Pythagorean Identity $sin^2\theta+cos^2\theta=1$ keeping our parameters in mind:

$sin^2\theta+cos^2\theta=1\\\\(-\frac{5}{13})^2+cos^2\theta=1\\ \\\frac{25}{169}+cos^2\theta=1\\ \\cos^2\theta=\frac{144}{169}\\ \\cos\theta=-\frac{12}{13}$

Thus, $sin2\theta=2sin\theta cos\theta=2(-\frac{5}{13})(-\frac{12}{13})=2(\frac{60}{169})=\frac{120}{169}$, which means the third option is correct.

Answer 2

60 is the answer I hope this is right