If $$t = \tan \frac{x}{2},$$ find expressions for $\sin x, \cos x$ in terms of $t$. Hence, solve the equation $$3\sin x - 4\cos x = 2.$$


I have been solving a lot of trig questions lately but this is different. I don't know how to approach this. I'm thinking of getting $\sin x$ and $\cos x$ from $\tan x$ and replacing in the equation but not sure how because of the $t$. Help please.

  • $\begingroup$ if you find any of these answers useful, feel free to upvote as many as youi wish. You can even accept the answer that you feel was most helpful (using the checkmark next to the answer). $\endgroup$ – robjohn Oct 9 '15 at 0:19

Indicated Solution

We can derive the Weierstrass Substitution:

Using the tangent double angle formula: $$ \tan(x)=\frac{2t}{1-t^2}\tag{1} $$ Then writing $\sec^2(x)$ in terms of $\tan^2(x)$ $$ \begin{align} \sec^2(x) &=1+\tan^2(x)\\ &=1+\frac{4t^2}{1-2t^2+t^4}\\ &=\frac{1+2t^2+t^4}{1-2t^2+t^4}\\ &=\left(\frac{1+t^2}{1-t^2}\right)^2\tag{2} \end{align} $$ Therefore, checking sign of $\cos(x)$ vs $\tan(x/2)$: $$ \cos(x)=\frac{1-t^2}{1+t^2}\tag{3} $$ Multiplying $(1)$ and $(3)$ gives $$ \sin(x)=\frac{2t}{1+t^2}\tag{4} $$ Then, as mentioned in comments, we simply need to solve for $t=\tan(x/2)$: $$ 3\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}-4\,\overbrace{\frac{1-t^2}{1+t^2}}^{‌​\cos(x)}=2\tag{5} $$ which is simply a quadratic equation in $t$ giving $$ \tan(x/2)=t=\frac{-3\pm\sqrt{21}}2\tag{6} $$

Alternate Solution

Suppose $\theta$ is an angle so that $\sin(\theta)=\frac45$ and $\cos(\theta)=\frac35$; that is, $\theta=\sin^{-1}\!\left(\frac45\right)$.

Then, $$ \begin{align} \sin(x-\theta) &=\cos(\theta)\sin(x)-\sin(\theta)\cos(x)\\[3pt] &=\frac35\sin(x)-\frac45\cos(x)\\ &=\frac25 \end{align} $$ which gives the two solutions $$ x=\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right) \implies\tan(x/2)=\frac{-3+\sqrt{21}}2 $$ and $$ x=\pi-\sin^{-1}\!\left(\frac25\right)+\sin^{-1}\!\left(\frac45\right) \implies\tan(x/2)=\frac{-3-\sqrt{21}}2 $$

  • $\begingroup$ Thank you. So what's the value of 3 sin x - 4 cos x = 2 from this. Should I just replace the value of sin x and cos x? $\endgroup$ – user274246 Oct 7 '15 at 9:18
  • $\begingroup$ Yes. $$3\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}-4\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)}=2$$ This is a quadratic equation in $t=\tan(x/2)$. $\endgroup$ – robjohn Oct 7 '15 at 9:24
  • $\begingroup$ Thanks. I got t= 1 or t= -3. That's right, right? $\endgroup$ – user274246 Oct 7 '15 at 9:33
  • $\begingroup$ I don't think that is right. Simplifying the equation I have above gives $t^2+3t-3=0$. $\endgroup$ – robjohn Oct 7 '15 at 9:48
  • $\begingroup$ Okay. Thank you. I get it now. Appreciate the help. $\endgroup$ – user274246 Oct 7 '15 at 10:04

Hint Rearranging gives (for $\frac{x}{2}$ in the image $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ of $\arctan$) that $$x = 2 \arctan t$$ (this is the celebrated if perhaps misnamed Weierstraß substitution, which has the convenient property of transforming rational functions in trigonometric functions of $x$ into rational functions in $t$, and which is particularly useful in computing antiderivatives of the former).

With this in hand, we can exploit the double-angle identity $\sin 2 u = 2 \sin u \cos u$ to write, e.g., $\sin x$ in terms of $t$: $$\sin x = \sin 2 \arctan t = 2 \sin \arctan t \cos \arctan t .$$ Appealing to a reference triangle (appropriately labeling a right triangle with legs $t$ and $1$ and acute angle $x$, computing the length of the remaining side, and using the definition of the trigonometric functions) gives us (algebraic) simplifications: $$\left\{ \begin{array}{rcl} \sin \arctan t &=& \frac{t}{\sqrt{1 + t^2}} \\ \cos \arctan t &=& \frac{1}{\sqrt{1 + t^2}}\end{array} \right.$$ Substituting gives $$\sin x = 2 \left(\frac{t}{\sqrt{1 + t^2}}\right) \left(\frac{1}{\sqrt{1 + t^2}}\right) = \frac{2t}{1 + t^2}.$$ One can similarly derive a rational expression in $t$ for $\cos x$ and hence write the given equation in $x$ as a rational equation in $t$.

(NB as hinted above, this only detects the solutions with $\frac{x}{2}$ in an appropriate range. To find all of the solutions, one must use these in conjunction with the usual symmetries of the trigonometric functions themselves.)


Hint: $\sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$, $\cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$


I am posting only because I don't see the most obvious approach:

Since $t=\tan(x/2)=\frac{\sin(x/2)}{\cos(x/2)}$ and $\sin^2(x/2)+\cos^2(x/2)=1,$ solve for $\sin(x/2)=\frac{t}{\sqrt{1+t^2}}$ and $\cos(x/2)=\frac{1}{\sqrt{1+t^2}}$ (here I am assuming that $x/2\in [0,\pi/2)$-the other cases are similar).

Now you can use the formulas that give you the $\sin$ and $\cos$ of double angles, and you are done.


If you know the bisection formulas $$ \left|\sin\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{2}} \qquad \left|\cos\frac{x}{2}\right|=\sqrt{\frac{1+\cos x}{2}} $$ you can put them together finding \begin{align} \left|\tan\frac{x}{2}\right|=\sqrt{\frac{1-\cos x}{1+\cos x}} &=\sqrt{\frac{1-\cos x}{1+\cos x}\frac{1-\cos x}{1-\cos x}}\\[6px] &=\sqrt{\frac{(1-\cos x)^2}{\sin^2x}}\\[6px] &=\biggl|\frac{1-\cos x}{\sin x}\biggr|\\[6px] &=\frac{1-\cos x}{|\sin x|} \end{align} It's easy to check that $\tan(x/2)$ has the same sign as $\sin x$ for every $x$ where $\sin x\ne0$ (and so $\tan(x/2)$ is defined), which allows us to remove the absolute values: $$ \tan\frac{x}{2}=\frac{1-\cos x}{\sin x} $$

We can do similarly by using $1+\cos x$ in the second step instead of $1-\cos x$, which gives another neat formula: $$ \tan\frac{x}{2}=\frac{\sin x}{1+\cos x} $$ Setting, for simplicity, $t=\tan(x/2)$, $X=\cos x$ and $Y=\sin x$, we can rewrite the equations as $$ \begin{cases} X+tY=1\\[6px] tX-Y=-1 \end{cases} $$ If we solve this for $X$ and $Y$, we arrive at $$ X=\cos x=\frac{1-t^2}{1+t^2},\qquad Y=\sin x=\frac{2t}{1+t^2} $$

If you apply these to your equation, you get $$ \frac{6t}{1+t^2}-\frac{4-4t^2}{1+t^2}=2 $$ that becomes a quadratic by removing the denominators: $$ 6t-4+4t^2=2+2t^2 $$ or, simplifying, $$ t^2+3t-3=0 $$

Note there's a small issue: you have to verify that the values of $x$ for which $\tan(x/2)$ is not defined are or are not solutions of the equations. The check is easily done: $\tan(x/2)$ is not defined for $x=k\pi$, but $$ 3\sin(k\pi)-4\cos(k\pi)=\pm4\ne2 $$ If the equation is $3\sin x-4\cos x=4$, you get a family of solutions not covered by the $t$-substitution.


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