Transforming a combination of sine and cosine functions to either sine or cosine function

Question

Consider the expression $a\sin\theta+b\cos\theta$. This expression consists of a combination of sine and cosine trigonometric functions which makes analysing it difficult. How can this expression be converted to an expression consisting of only sine or cosine functions?

Answer 1

Suppose $y=a\sin\theta \pm b\cos\theta$

In many cases, an expression of this type, consisting of a combination of sine and cosine functions is difficult to analyze. So we try to convert the entire expression into an expression consisting of only one trigonometric function, that is, either sine or cosine function. Here the thought process and the process for converting it into an expression consisting of only sine function is explained. The process for converting it into an expression consisting of only cosine function is similar.

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We know that $\sin(x \pm y)=\sin x\cos y \pm \cos x\sin y$

We notice that there are two terms in the formula for $\sin(x \pm y)$, $\sin x\cos y$ and $\cos x\sin y$. We also notice that there are two instances in which the sum of squares of factors one from each of two terms add up to 1.

Instance 1: Let's take $\sin x$ from $\sin x \cos y$ and $\cos x$ from $\cos x \sin y$. Squaring and adding both the factors we get, $\sin^2 x+\cos^2 x$ which is equal to one.

Instance 2: Let's take $\cos y$ from $\sin x \cos y$ and $\sin y $ from $\cos x \sin y$. Squaring and adding both the factors we get, $\cos^2 y+\sin^2 y$ which is equal to one.

This can be regarded as the condition for an expression of the type $y=a\sin\theta \pm b\cos\theta$ to be converted into an expression consisting of only sine function.

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So if we want to convert an expression of the type $y=a\sin\theta \pm b\cos\theta$ to an expression consisting of only sine function, we have to first check if it satisfies this condition.

Let's take $\sin\theta$ from $a\sin\theta$ and $\cos\theta$ from $b\cos\theta$. Squaring and adding both the factors we get, $\sin^2\theta+\cos^2\theta$ which is equal to one.

Let's take $a$ from $a\sin\theta$ and $b$ from $b\cos\theta$.Squaring and adding both the factors we get $a^2+b^2$. We do not know whether $a^2+b^2$ is 1 or not. We need to transform both in such a way that the sum of their squares is 1. Let us multiply $a$ and $b$ by a number $k$ such that $(ka)^2+(kb)^2=1$

$\implies k^2 a^2+k^2 b^2=1$

$\implies k^2(a^2+b^2)=1$

$\implies k^2=\frac{1}{a^2+b^2}$

$\implies k=\frac{1}{\sqrt{a^2+b^2}}$

If $a$ and $b$ are multiplied by $k$, the sum of their squares is 1. But if we multiply $a$ and $b$ by $k$ then we also have to divide $a$ and $b$ by $k$ so that their original values remain unaffected.

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$y=a\sin\theta \pm b\cos\theta$

$\implies y=\frac{ka\sin\theta}{k} \pm \frac{kb\cos\theta}{k}$

$\implies y=\frac{1}{k}(ka\sin\theta \pm kb\cos\theta)$

$\implies y=\frac{1}{\frac{1}{\sqrt{a^2+b^2}}}(\frac{a}{\sqrt{a^2+b^2}}\sin\theta \pm \frac{b}{\sqrt{a^2+b^2}}\cos\theta)$

$\implies y=\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin\theta \pm \frac{b}{\sqrt{a^2+b^2}}\cos\theta)$

Let $\frac{a}{\sqrt{a^2+b^2}}=\cos\alpha$

Then, $\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\frac{a^2}{a^2+b^2}}=\sqrt{\frac{a^2+b^2-a^2}{a^2+b^2}}=\frac{b}{\sqrt{a^2+b^2}}$

$\therefore y=\sqrt{a^2+b^2}(\cos\alpha \sin\theta \pm \sin\alpha \cos \theta)=\sqrt{a^2+b^2}(\sin\theta\cos\alpha \pm \cos \theta \sin\alpha)=\sqrt{a^2+b^2}\sin(\theta \pm \alpha)$