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National Book Foundation (NBF) Class 11 Math Exercise 8.1 Solution

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First Year Math New Book Exercise 8.1 Solution

To solve the trigonometric identities for given angles:

  1. Find the values of cos⁡(α±β)\cos(\alpha \pm \beta), sin⁡(α±β)\sin(\alpha \pm \beta), tan⁡(α±β)\tan(\alpha \pm \beta) for each given pair of angles.

    (i) Given: α=180∘\alpha = 180^\circ, β=60∘\beta = 60^\circ

    Solution:

    cos⁡(α+β)=cos⁡(180∘+60∘)=−cos⁡60∘=−12cos⁡(α−β)=cos⁡(180∘−60∘)=−cos⁡120∘=−12sin⁡(α+β)=sin⁡(180∘+60∘)=−sin⁡60∘=−32sin⁡(α−β)=sin⁡(180∘−60∘)=sin⁡60∘=32tan⁡(α+β)=tan⁡(180∘+60∘)=tan⁡60∘=3tan⁡(α−β)=tan⁡(180∘−60∘)=−tan⁡60∘=−3\begin{align*} \cos(\alpha + \beta) &= \cos(180^\circ + 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \\ \cos(\alpha - \beta) &= \cos(180^\circ - 60^\circ) = -\cos 120^\circ = -\frac{1}{2} \\ \sin(\alpha + \beta) &= \sin(180^\circ + 60^\circ) = -\sin 60^\circ = -\frac{\sqrt{3}}{2} \\ \sin(\alpha - \beta) &= \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \\ \tan(\alpha + \beta) &= \tan(180^\circ + 60^\circ) = \tan 60^\circ = \sqrt{3} \\ \tan(\alpha - \beta) &= \tan(180^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3} \end{align*}

    (ii) Given: α=90∘\alpha = 90^\circ, β=45∘\beta = 45^\circ

    Solution:

    cos⁡(α+β)=cos⁡(90∘+45∘)=cos⁡135∘=−22cos⁡(α−β)=cos⁡(90∘−45∘)=cos⁡45∘=22sin⁡(α+β)=sin⁡(90∘+45∘)=sin⁡135∘=22sin⁡(α−β)=sin⁡(90∘−45∘)=sin⁡45∘=22tan⁡(α+β)=tan⁡(90∘+45∘)=−cot⁡45∘=−1tan⁡(α−β)=tan⁡(90∘−45∘)=tan⁡45∘=1 \begin{align*} \cos(\alpha + \beta) &= \cos(90^\circ + 45^\circ) = \cos 135^\circ = -\frac{\sqrt{2}}{2} \\ \cos(\alpha - \beta) &= \cos(90^\circ - 45^\circ) = \cos 45^\circ = \frac{\sqrt{2}}{2} \\ \sin(\alpha + \beta) &= \sin(90^\circ + 45^\circ) = \sin 135^\circ = \frac{\sqrt{2}}{2} \\ \sin(\alpha - \beta) &= \sin(90^\circ - 45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2} \\ \tan(\alpha + \beta) &= \tan(90^\circ + 45^\circ) = -\cot 45^\circ = -1 \\ \tan(\alpha - \beta) &= \tan(90^\circ - 45^\circ) = \tan 45^\circ = 1 \end{align*}

    (iii) Given: α=180∘\alpha = 180^\circ, β=30∘\beta = 30^\circ

    Solution:

    cos⁡(α+β)=cos⁡(180∘+30∘)=−cos⁡30∘=−32cos⁡(α−β)=cos⁡(180∘−30∘)=−cos⁡30∘=−32sin⁡(α+β)=sin⁡(180∘+30∘)=−sin⁡30∘=−12sin⁡(α−β)=sin⁡(180∘−30∘)=sin⁡30∘=12tan⁡(α+β)=tan⁡(180∘+30∘)=−tan⁡30∘=−13tan⁡(α−β)=tan⁡(180∘−30∘)=−tan⁡30∘=−13\begin{align*} \cos(\alpha + \beta) &= \cos(180^\circ + 30^\circ) &= -\cos 30^\circ = -\frac{\sqrt{3}}{2} \\[10pt] \cos(\alpha - \beta) &= \cos(180^\circ - 30^\circ) &= -\cos 30^\circ = -\frac{\sqrt{3}}{2} \\[10pt] \sin(\alpha + \beta) &= \sin(180^\circ + 30^\circ) &= -\sin 30^\circ = -\frac{1}{2} \\[10pt] \sin(\alpha - \beta) &= \sin(180^\circ - 30^\circ) &= \sin 30^\circ = \frac{1}{2} \\[10pt] \tan(\alpha + \beta) &= \tan(180^\circ + 30^\circ) &= -\tan 30^\circ = -\frac{1}{\sqrt{3}} \\[10pt] \tan(\alpha - \beta) &= \tan(180^\circ - 30^\circ) &= -\tan 30^\circ = -\frac{1}{\sqrt{3}} \end{align*}
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