# find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$

If $$\sin\theta=\frac{3}{4}$$ and $$\tan\theta=\frac{9}{2}$$ then find $$\cos\theta$$

The solution is given in my reference as $$\frac{1}{6}$$.

$$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}$$ $$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\sqrt{1+\tan^2\theta}}=\frac{1}{\sqrt{1+\frac{81}{4}}}=\frac{2}{\sqrt{85}}$$ $$\cos\theta=\frac{\sin\theta}{\tan\theta}=\frac{\frac{3}{4}}{\frac{9}{2}}=\frac{3}{4}.\frac{2}{9}=\frac{1}{6}$$ Why is it getting confused here ?

• – Mohammad Zuhair Khan Oct 19 '18 at 13:17
• It is confusing because no $\theta$ could satisfies both of the assumptions. – xbh Oct 19 '18 at 13:21

Because you have two different problems. Obviously (from your calculations which are correct) it can't be $$\sin\theta=\frac{3}{4}$$ and $$\tan\theta=\frac{9}{2}$$ at the same time.

• You mean "can't", not "can" right? – projectilemotion Oct 19 '18 at 13:08
• Surely you mean "it cannot be"! – user247327 Oct 19 '18 at 13:08
• well. so u saying the actual question in my reference is wrong, right ? – ss1729 Oct 19 '18 at 13:10
• Yes, that is what I'm saying. – Aqua Oct 19 '18 at 13:10

This is an impossible situation. If you use the identity $$\tan(\theta) = {\sin(\theta)\over\cos(\theta)},$$ you get $$\cos(\theta) = 1/6$$.

However, $$\cos(\theta) = \sqrt{7}/4$$ by the Pythagorean identity.

The problem is actually incorrect. Here is why:

Consider a right triangle and angle $$\theta$$ in that triangle such that $$\sin\theta=\dfrac{3}{4}.$$

That means that the right triangle has one leg $$3$$ and hypotenuse $$4$$. This implies that the other leg is $$\sqrt{4^2-3^2}=\sqrt{7}.$$

This implies that $$\cos \theta=\dfrac{\sqrt{7}}4.$$ Now also note that $$\tan \theta \neq \frac 92$$ as stated in your problem.