Well, I know that this is not a homework help site. This isn't a homework, but rather I want to clear a probable misconception that I have on trigonometry submultiple and multiple angles. Here is the question.

Determine the smallest positive value of x(in degrees) for which $tan (x+100°)=tan (x+50).tan x. tan(x-50)$

I proceeded in the following way. $\frac {tan (x+100)}{tan (x-50)} = tan(x+50)tan x$ $\frac{sin(x+100)cos(x-50)}{cos(x+100)sin(x-50)}=\frac{sin(x+50).sin x}{cos(x+50).cos x}$. Doing componendo-dividendo on both sides, and simplifying them in the form of compound angles I am getting, $\frac{sin (2x+50)}{sin 150}=\frac{cos 50}{-cos(2x+50)}.$ Cross multiplying, and multiplying 2 on both sides, and on LHS, using the multiple angle formula ,I get, $sin (4x+100)=-2 sin 150. cos 50$ $\implies sin (4x°+100°)=-cos 50° (sin 150= cos 60= 1/2)$ $\implies sin(4x+100°)=-sin(90°+50°)$ $\implies 4x+100=140$ $\implies x=10$ But the answer is given x=30. It only comes if in step 1, I use the following step, $sin (4x+100)=sin (270-50)$ But then, if $smallest$ positive value is required, why I will chose the third quadrant instead of second quadrant? Is the answer wrong? Or I am having some misconceptions? Is my procedure correct?


everything seems fine until you get to the line $$\sin(4x+100)=-\cos 50$$which you now equate to $$\sin(90+50)$$ which is not correct because $$\sin(90+50)=\sin90\cos 50+\cos 90\sin50=\cos 50$$

Instead, use $$-\cos 50=\cos(180-50)=\cos130=\sin(90-130)=\sin(-40)$$

Then you have $$4x+100=\begin{cases}-40+n.360 \\180+40+n.360\end{cases}$$

  • $\begingroup$ Why? Isn't this associated angles working here? $\endgroup$ – Aneek Sep 9 '15 at 11:29
  • $\begingroup$ $sin (90+\theta)= -cos \theta$. Isn't it? $\endgroup$ – Aneek Sep 9 '15 at 11:30
  • $\begingroup$ $\sin(90+\theta)=\sin90\cos\theta+\cos90\sin\theta$ $\endgroup$ – David Quinn Sep 9 '15 at 11:33
  • $\begingroup$ So sin 90=1, cos 90=0, so you get $sin (90+\theta)=cos\theta$ $\endgroup$ – Aneek Sep 9 '15 at 11:34
  • $\begingroup$ yes I agree.... $\endgroup$ – David Quinn Sep 9 '15 at 11:56

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