CBSE 2025 · Region 5 · Set 1 · Q23 · 2 marks
It is given that $\displaystyle \sin (\mathrm{A}-\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}$. Use it to find the value of $\displaystyle \sin 15^{\circ}$.If $\displaystyle \sin \mathrm{A}=\mathrm{y}$, then express $\displaystyle \cos \mathrm{A}$ and $\displaystyle \tan \mathrm{A}$ in terms of y .
It is given that $\displaystyle \sin (\mathrm{A}-\mathrm{B})=\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}$. Use it to find the value of $\displaystyle \sin 15^{\circ}$.
If $\displaystyle \sin \mathrm{A}=\mathrm{y}$, then express $\displaystyle \cos \mathrm{A}$ and $\displaystyle \tan \mathrm{A}$ in terms of y .
Marking-scheme solution
\[\begin{aligned}
\sin 15^{\circ} & =\sin \left(45^{\circ}-30^{\circ}\right) \\
& =\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ} \\
& =\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2} \\
& =\frac{\sqrt{3}-1}{2 \sqrt{2}} \text { or } \frac{\sqrt{6}-\sqrt{2}}{4}
\end{aligned}
\]
\(\displaystyle \cos \mathrm{A}=\sqrt{1-\sin ^{2} \mathrm{~A}}=\sqrt{1-\mathrm{y}^{2}}\)
\(\displaystyle \tan \mathrm{A}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\frac{\mathrm{y}}{\sqrt{1-\mathrm{y}^{2}}}\)
Introduction to TrigonometryTrigonometric Ratios of Some Specific AnglesApplyvery_short_answermedium
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.