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These are a few trig identities questions that I just can't figure out. They are from the Cambridge 3U book.

I hate to put more than one question up at a time but I just can't figure any of them out, anyhow:

  1. Eliminate theta from the following pair of equations:

\begin{align}x &= \sin(\theta) - 3\cos(\theta)\\ y &= \sin(\theta) + 2\cos(\theta)\end{align}

  1. If $\tan(\theta) + \sin(\theta) = x$ and $\tan(\theta) - \sin(\theta) = y$, prove that $$x^4 + y^4 = 2xy(8 + xy)$$

  2. If $\dfrac a{\sin A}= \dfrac b{\cos A}$, show that $$\sin(A)\cos(A) = \frac{ab}{a^2 + b^2}$$

  3. If $\dfrac{a + b}{\text{cosec}(x)} = \dfrac{a - b}{\cot(x)}$, show that $$\text{cosec}(x)\cot(x) = \frac{a^2 - b^2}{4ab}$$

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    $\begingroup$ Please use MathJax for a better reading and avoid multiple questions in one post. More information about MathJax: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – jacmeird Jul 10 '17 at 7:58
  • $\begingroup$ Thanks for the help, I think u messed up the last question though $\endgroup$ – Isaac Greene Jul 10 '17 at 8:11
  • $\begingroup$ Oh I'm sorry. I've corrected this with my phone. A bit tricky! $\endgroup$ – jacmeird Jul 10 '17 at 8:12
  • $\begingroup$ For (1) we have $\sin \theta=(2x+3y)/5$ and $\cos \theta=-(x+y).$ Therefore $1=\sin^2\theta+\cos^2\theta=(2x+3y)^2/25+(x+y)^2.$ $\endgroup$ – DanielWainfleet Jul 10 '17 at 10:22
  • $\begingroup$ For (3) let $a=x\sin A$ and $B=x\cos A.$ Then $ab=x^2\sin A \cos A$ and $a^2+b^2=x^2(\sin^2A+\cos^2A)=x^2.$ $\endgroup$ – DanielWainfleet Jul 10 '17 at 10:30
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  1. Put $a = \sin\theta$ and $b = \cos\theta$. Two equations in two variables $a$ and $b$. Solve for $a$ and $b$ and then use $a^2 + b^2 = 1$.

  2. Directly substitute $x$ and $y$ and check LHS and RHS.

  3. Square both sides and use $\sin^2A + \cos^2A = 1$.

  4. $\cos x$ is given. Using $\sin^2x + \cos^2x = 1$, compute $\sin x$ and then substitute in last equation.

There might be other simpler ways to solve these. But for now these methods will keep you going.

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$4)\dfrac{a+b}{\csc x}=\dfrac{a-b}{\cot x}=\pm\sqrt{\dfrac{(a+b)^2-(a-b)^2}{\csc^2x-\cot^2x}}=?$

Observe that $\csc x,\cot x$ will have the same sign.

$(3)$ Can you use the method applied in $(4)$

$(2)$ Solve for $\tan\theta,\sin\theta$

Use $$\dfrac1{\sin^2\theta}-\dfrac1{\tan^2\theta}=\cdots=1$$

$(1)$ Solve for $\sin\theta,\cos\theta$

Use Fundamental Theorem of Trigonometry

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  • $\begingroup$ Could you explain how you came to the large square root in the second line please $\endgroup$ – Isaac Greene Jul 10 '17 at 8:22
  • $\begingroup$ @Iso1234, If $$\dfrac px=\dfrac qy$$ $=k$(say) $$\sqrt{\dfrac{p^2-q^2}{x^2-y^2}}=?$$ assuming $k\ne\pm1$ $\endgroup$ – lab bhattacharjee Jul 10 '17 at 8:28
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Multiply the two fractions together the answer will be the the square of any of them , sin^2A+cos^2A=1 ab/(sinAcosA)=a^2/sin^2A=b^2/cos^2A=a^2+b^2/1 Then make cross multiplication to get the answer

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  • $\begingroup$ For 4- you will do the same steps except you will subtract instead of addition $\endgroup$ – Ahmed Nabih Jul 14 '17 at 0:11

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