Porisma

Worked solution

Control the square-root sign in a trigonometric substitution

If x=2tan⁡θx=2\tan\theta, where 0∘<θ<90∘0^\circ<\theta<90^\circ, show that x4+x2=sin⁡θ\dfrac{x}{\sqrt{4+x^2}}=\sin\theta.

CJC · O Level revision · Example 19

Answer

x4+x2=sin⁡θ\dfrac{x}{\sqrt{4+x^2}}=\sin\theta.

Compare your reasoning

Substitute and factor the radicand

x4+x2=2tan⁡θ4+(2tan⁡θ)2=2tan⁡θ4(1+tan⁡2θ)\begin{aligned}&\frac{x}{\sqrt{4+x^2}}\\ & =\frac{2\tan\theta}{\sqrt{4+(2\tan\theta)^2}}\\ & =\frac{2\tan\theta}{\sqrt{4(1+\tan^2\theta)}}\end{aligned}

Retain the source left-side chain before applying the basic identity.

Use the acute-angle sign

2tan⁡θ4sec⁡2θ=2tan⁡θ2sec⁡θ\frac{2\tan\theta}{\sqrt{4\sec^2\theta}}=\frac{2\tan\theta}{2\sec\theta}

The source replaces 1+tan⁡2θ1+\tan^2\theta with sec⁡2θ\sec^2\theta. Independently make the sign step explicit: the square root is 2∣sec⁡θ∣2|\sec\theta|, and the given acute interval makes secθ positive.

Convert the quotient to sine and cosine

tan⁡θsec⁡θ=sin⁡θcos⁡θcos⁡θ1=sin⁡θ\frac{\tan\theta}{\sec\theta}=\frac{\sin\theta}{\cos\theta}\frac{\cos\theta}{1}=\sin\theta

Both denominator factors are nonzero on the given interval.

How did this question go?

Common mistakes

Outside the given acute interval, sec⁡2θ=∣sec⁡θ∣\sqrt{\sec^2\theta}=|\sec\theta|; dropping the absolute value can change the sign.