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 $$\text{cosec}(x)\cot(x) = \frac{a^2 - b^2}{4ab}$$

  • 3
    $\begingroup$ 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
  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.


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

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

  • $\begingroup$ 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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.