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Question: How can \(\tan 4x\) be simplified or \(\sec 2x\) ?...

How can tan⁡4x\tan 4x be simplified or sec⁡2x\sec 2x ?

Explanation

Solution

In the given question, we are given two trigonometric functions and we have to simplify them. By simplifying, we mean that we have to replace the given function with some other known value of the function until it cannot be done further, that is, until there is no simpler value of the given function. The given trigonometric functions have large angles and we know several trigonometric identities to convert large-angle functions into smaller ones so with the help of those values we have to simplify the given functions.

Complete step by step answer:
We know that –
tan⁡x=sin⁡xcos⁡x ⇒tan⁡4x=sin⁡4xcos⁡4x  \tan x = \dfrac{{\sin x}}{{\cos x}} \\\ \Rightarrow \tan 4x = \dfrac{{\sin 4x}}{{\cos 4x}} \\\
We also know that –
sin⁡2x=2sin⁡xcos⁡x and cos⁡2x=cos⁡2x−sin⁡2x ⇒sin⁡4x=2sin⁡2xcos⁡2x and cos⁡4x=cos⁡22x−sin⁡22x  \sin 2x = 2\sin x\cos x\,and\,\cos 2x = {\cos ^2}x - {\sin ^2}x \\\ \Rightarrow \sin 4x = 2\sin 2x\cos 2x\,and\,\cos 4x = {\cos ^2}2x - {\sin ^2}2x \\\
Using these values in the obtained equation, we get –
tan⁡4x=2sin⁡2xcos⁡2xcos⁡22x−sin⁡22x ⇒tan⁡4x=2×2sin⁡xcos⁡x(cos⁡2x−sin⁡2x)(cos⁡2−sin⁡2x)−(2sin⁡xcos⁡x)2 ⇒tan⁡4x=4sin⁡xcos⁡x(1−sin⁡2x−sin⁡2x)(1−sin⁡2x−sin⁡2x)−4sin⁡2xcos⁡2x ⇒tan⁡4x=4sin⁡xcos⁡x(1−2sin⁡2x)(1−2sin⁡2x)−4sin⁡2xcos⁡2x  \tan 4x = \dfrac{{2\sin 2x\cos 2x}}{{{{\cos }^2}2x - {{\sin }^2}2x}} \\\ \Rightarrow \tan 4x = \dfrac{{2 \times 2\sin x\cos x({{\cos }^2}x - {{\sin }^2}x)}}{{({{\cos }^2} - {{\sin }^2}x) - {{(2\sin x\cos x)}^2}}} \\\ \Rightarrow \tan 4x = \dfrac{{4\sin x\cos x(1 - {{\sin }^2}x - {{\sin }^2}x)}}{{(1 - {{\sin }^2}x - {{\sin }^2}x) - 4{{\sin }^2}x{{\cos }^2}x}} \\\ \Rightarrow \tan 4x = \dfrac{{4\sin x\cos x(1 - 2{{\sin }^2}x)}}{{(1 - 2{{\sin }^2}x) - 4{{\sin }^2}x{{\cos }^2}x}} \\\
The above-obtained equation cannot be simplified further.
Now,
sec⁡2x=1cos⁡2x ⇒sec⁡2x=1cos⁡2x−sin⁡2x ⇒sec⁡2x=11−2sin⁡2x  \sec 2x = \dfrac{1}{{\cos 2x}} \\\ \Rightarrow \sec 2x = \dfrac{1}{{{{\cos }^2}x - {{\sin }^2}x}} \\\ \Rightarrow \sec 2x = \dfrac{1}{{1 - 2{{\sin }^2}x}} \\\
Hence, the simplified form of tan⁡4x\tan 4x is 4sin⁡xcos⁡x(1−2sin⁡2x)(1−2sin⁡2x)\-4sin⁡2xcos⁡2x\dfrac{{4\sin x\cos x(1 - 2{{\sin }^2}x)}}{{(1 - 2{{\sin }^2}x) \- 4{{\sin }^2}x{{\cos }^2}x}} and that of sec⁡2x\sec 2x is 11−2sin⁡2x\dfrac{1}{{1 - 2{{\sin }^2}x}} .

Note: The ratio of two sides of a right-angled triangle is known as the trigonometric ratios. Trigonometry consists of sine, cosine, tangent, secant, cosecant and cotangent functions. Sine, cosine and tangent are the main three function while cosecant, secant and cotangent functions are their reciprocals respectively, thus one trigonometric ratio can be converted into the trigonometric ratio of another function by using this knowledge or the trigonometric identities like we have used a few identities in this solution to convert the tangent function into the sine and cosine terms and then sine and cosine terms into smaller terms. At last, we obtained the trigonometric functions in terms of x, which is the simplest form and cannot be solved further by any identity.