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Find the value of $$\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+...... +\cos^2(\theta+179^{\circ})$$

Attempt, $$\cos x=-\cos(180^\circ-x),\sin x=\cos(90^\circ-x),\cos x=\sin(90^\circ-x),\sin x=\sin(180^\circ-x)$$

$$\cos^2\theta+\cos^2(\theta+1^\circ)+\cos^2(\theta+ 2^\circ)+ \dots+\cos^2(\theta+179^\circ)=\cos^2\theta+\sum_{ n=1}^{179}\cos^2(\theta+n^\circ)$$

$$\cos^2(\theta+1^\circ)=\cos^2\theta\cos^21^\circ-2\cos\theta\cos1^\circ\sin\theta\sin1^\circ+\sin^2 \theta\sin^21^\circ$$

$$\cos^2(\theta+179^\circ)=\cos^2\theta\cos^2179^ \circ-2\cos\theta\cos179^\circ\sin\theta\sin179^\circ+ \sin^2\theta \sin^2179^\circ=\cos^2\theta\cos^2179^\circ+2\cos \theta\cos1^\circ \sin\theta\sin1^\circ+\sin^2 \theta\sin^2179^\circ$$

$$\cos^2\theta+\sum_{n=1}^{179}\cos^2(\theta+n^\circ )=\cos^2\theta+\sum_{n=1}^{179}(\cos^2\theta\cos^2 n^\circ+\sin^2\theta \sin^2n^\circ)$$

$$=\cos^2\theta+\sin^2\theta+2\sum_{n=1}^{89}(\cos^2 \theta\cos^2n^\circ+\sin^2\theta\sin^2n^\circ)$$

$$=1+2\sum_{n=1}^{89}\cos^2n^\circ=1+2\left(44+ \dfrac12 \right)=90$$

Mathxx
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3 Answers3

8

Let $$f(\theta)=\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+...... +\cos^2(\theta+179^{\circ})$$

Then $$f(\theta+1)-f(\theta)=\cos^2(\theta+180^{\circ})-\cos ^2\theta =0$$

This shows that $f(\theta)$ is periodic with period $1$. In particular $$f(\theta)=f(\theta+90)$$

Show now that $$f(\theta+90)=\sin ^2\theta+\sin^2 (\theta+1^{\circ})+\sin^2(\theta+2^{\circ})+...... +\sin^2(\theta+179^{\circ})$$

and thus $$2 f(\theta)=f(\theta)+f(\theta+90)=180$$

N. S.
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$$\begin{align} \sum_{k=0}^{179}\cos^2(\theta+k\cdot 1^\circ) &= \sum_{k=0}^{89}\cos^2(\theta+k\cdot 1^\circ) + \cos^2(\theta+k\cdot 1^\circ + 90^\circ) \\ &= \sum_{k=0}^{89}\cos^2(\theta+k\cdot 1^\circ) + \sin^2(\theta+k\cdot 1^\circ)\\ &= \sum_{k=0}^{89} \; 1 \\ &= 90 \end{align}$$

Blue
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We have $$\cos^2(a+kb) = \dfrac{1+\cos(2a+2kb)}2 = \dfrac{1+\cos(2a)\cos(2kb)-\sin(2a)\sin(2kb)}2$$ Hence, we have \begin{align} \sum_{k=0}^n \cos^2(a+kb) & = \dfrac{n+1}2 + \dfrac{\cos(2a)}2\cdot \sum_{k=0}^n \cos(2kb)- \dfrac{\sin(2a)}2 \cdot \sum_{k=0}^n \sin(2kb) \end{align} Now from here, we have $\displaystyle\sum_{k=0}^n \cos(kt)$ and $\displaystyle\sum_{k=0}^n \sin(kt)$. Hence, we have \begin{align} \sum_{k=0}^n \cos^2(a+kb) & = \dfrac{n+1}2 + \left(\dfrac{\cos(2a)}2 \cdot \cos(nb) - \dfrac{\sin(2a)}2 \cdot \sin(nb)\right) \cdot \dfrac{\sin((n+1)b)}{\sin(b)}\\ & = \dfrac{n+1}2 + \dfrac{\cos(2a+nb)}2 \cdot \dfrac{\sin((n+1)b)}{\sin(b)} \end{align} In your case, $a=\theta$, $b=1^{\circ}$ and $n=179$. This gives us the sum to be $$\dfrac{179+1}2 + \dfrac{\cos((2\theta+179)^{\circ})}2 \cdot \dfrac{\sin(180^{\circ})}{\sin(1^{\circ})} = 90$$

Adhvaitha
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