Q.Find the value of the expression $\sin\left(2\tan^{-1}\frac{1}{3}\right)+\cos\left(\tan^{-1}2\sqrt2\right)$.

Concept: Inverse trigonometric functions and their composition with trigonometric ratios — using right-triangle geometry or substitution.

Let $\theta = \tan^{-1}\frac{1}{3}$. Then $\tan\theta = \frac{1}{3}$.

We need $\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\cdot\frac13}{1+\frac19} = \frac{2/3}{10/9} = \frac{3}{5}$.

Now let $\phi = \tan^{-1}2\sqrt2$. Then $\tan\phi = 2\sqrt2$. …

Concept & Intuition

When you see $\tan^{-1}\frac{1}{3}$, think: "this is an angle whose tangent is $\frac{1}{3}$." Draw a right triangle where the opposite side is $1$ and the adjacent side is $3$. The hypotenuse then comes from Pythagoras. Once you have all three sides, you can read off $\sin$ and $\cos$ of that angle directly — no calculator needed.

The same idea works for $\tan^{-1}2\sqrt2$. That's an angle whose tangent is $2\sqrt2$. Again, build the triangle: opposite $= 2\sqrt2$, adjacent $= 1$, find the hypotenuse.

Then the problem becomes just plugging into $\sin(2\theta)$ and $\cos(\phi)$ — both of which are straightforward with the triangle values.

Step-by-step solution

1. Handle $\sin\left(2\tan^{-1}\frac{1}{3}\right)$

Let $\theta = \tan^{-1}\frac{1}{3}$. Then $\tan\theta = \frac{1}{3}$ and $\theta$ is acute ($0 < \theta < \frac{\pi}{2}$).

Draw a right triangle with opposite $= 1$, adjacent $= 3$. Hypotenuse:

So:

Now use the double-angle identity:

$\sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \frac{1}{\sqrt{10}} \cdot \frac{3}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5}$ …

Method: Trig functions of inverse-tangent angles via a right triangle

To evaluate $\sin$, $\cos$ (or their multiple-angle versions) of an angle given as $\tan^{-1}k$, model the angle with a right triangle so every ratio can be read off directly.

Steps

Step 1: Name each inverse as an angle and build its triangle.

If $\theta=\tan^{-1}k$, take opposite $=k$, adjacent $=1$, so hypotenuse $=\sqrt{1+k^2}$. Then $\sin\theta=\tfrac{k}{\sqrt{1+k^2}}$, $\cos\theta=\tfrac{1}{\sqrt{1+k^2}}$.

Step 2: For a doubled angle, use a double-angle identity.

Step 3: Read the single-angle terms straight from the triangle. …

Common Mistakes

Mistake 1: Computing $\sin\!\big(2\tan^{-1}\tfrac13\big)$ as $2\sin\!\big(\tan^{-1}\tfrac13\big)$.

Why it's wrong: $\sin 2\theta\ne2\sin\theta$. Correct approach: use $\sin 2\theta=2\sin\theta\cos\theta=\tfrac{2t}{1+t^2}$ with $t=\tfrac13$, giving $\tfrac35$.

Mistake 2: Slipping the hypotenuse for $\tan^{-1}(2\sqrt2)$. …