I didn't do well on these problems on my exam. I was wondering if someone can guide me through these problems step by step.

  • Use identities to write the expression as a single function of $x$:


  • Verify identeties.

$\qquad (a) \quad \sin 4x=4\sin x \cos x \cos 2x$

$\qquad (b) \quad \tan x+\cot x=2\csc 2x$

$\qquad (c) \quad \sec^2 \dfrac{x}{2}=\dfrac{2}{1+\cos x}$

  • 1
    $\begingroup$ Unfortunately the site does not really work this way. You're expected to show a little more effort (like explaining what you already think about the problem, and trying to explain what in particular you're having trouble with) beyond just a problem statement or the question is likely to be closed and poorly-received. In addition, it rarely goes well when a question contains multiple unrelated problems. $\endgroup$ – pjs36 Apr 18 '17 at 3:18
  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Apr 18 '17 at 3:19
  • $\begingroup$ Try to post separate problems and at least show part of your attempts, it might help you since someone can tell you where you went wrong. $\endgroup$ – goldenlinx Apr 18 '17 at 3:32

As pjs36 said, you're expected to show some effort. Since you haven't left any, I will only give some tips.

  • You do have an identity for tangent of sums of angles. If you don't know about it, try to write $\tan(a+b)=\frac{sin(a+b)}{cos(a+b)}$ and work from there.
  • To verify identities:
    1. $\sin(4x)=\sin(2\cdot(2x))$
    2. $\tan(x)=\frac{\sin(x)}{\cos(x)}$, $\cot(x)=\frac{1}{\tan(x)}$
    3. $\cos(2x)=2\cos^2(x)-1$
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To answer one of your questions about the trigonometric identities, here is the proof. $$\sin 4x=4\sin x \cos x \cos 2x$$

I find in this case it is easier to start on the left side because it is easier to break apart an identity, rather than condense.

When looking at the various trigonometric identities, it is important to first find the identity that is most similar to the format of $\sin(4x)$. This would be: $$ \sin(2x) = 2 \sin(x) \cos(x) $$

Now using this identity we can change the LHS of the equation to be: $$ 2 \sin(2x) \cos(2x) $$

Now this is getting more similar to the RHS of the equation. From here we already see something we just used an identity for - $\sin(2x)$

$$ 2(2\sin(x)\cos(x))\cos(2x) $$


This brings you to LHS=RHS

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