Trigonometrical Ratios of Multiple & Sub-multiple Angles

Trigonometrical Ratios of Multiple & Sub-multiple Angles

Here again, we will start with a list of the formulae regarding the present topic Trigonometrical Ratios of Multiple & Sub-multiple Angles. Skip to the examples here. Alternatively, you may choose to start over with the Measurement of Angles or go through the previous post on Trigonometrical Ratios of the Sum and Difference of Two Angles.

Formulae

\[
\sin(2A) = 2\sin(A)\cos(A)
\]\[
\cos(2A) = \cos^2(A) – \sin^2(A) = 1 – 2\sin^2A = 2\cos^2(A) – 1
\]\[
\tan(2A) = \frac{2\tan(A)}{1 – \tan^2(A)}
\]\[
\sin(3A) = 3\sin(A) – 4\sin^3A
\]\[
\cos(3A) = 4\cos^3(A) – 3\cos(A)
\]\[
\tan(3A) = \frac{3\tan(A) – \tan^3(A)}{1 – \tan^2(A)}
\]\[
\sin(A) = 2\sin\mkern-5mu\left( \frac{A}{2} \right)\cos\mkern-5mu\left( \frac{A}{2} \right)
\]\[
\cos(A) = \cos^2\mkern-5mu\left( \frac{A}{2} \right) – \sin^2\mkern-5mu\left( \frac{A}{2} \right) = 2\cos^2\mkern-5mu\left( \frac{A}{2} \right) – 1 = 1 – 2\sin^2\mkern-5mu\left( \frac{A}{2} \right)
\]\[
\tan(A) = \frac{2\tan\mkern-5mu\left( \frac{A}{2} \right)}{1 – \tan^2\mkern-5mu\left( \frac{A}{2} \right)}
\]\[
\sin\mkern-5mu\left( \frac{A}{2} \right) = \pm \sqrt{\frac{1 – \cos(A)}{2}} = \frac{\pm \sqrt{1 + \sin(A)} \pm \sqrt{1 – \sin(A)}}{2}
\]\[
\cos\mkern-5mu\left( \frac{A}{2} \right) = \pm \sqrt{\frac{1 + \cos(A)}{2}} = \frac{\pm \sqrt{1 + \sin(A)} \mp \sqrt{1 – \sin(A)}}{2}
\]\[
\tan\mkern-5mu\left( \frac{A}{2} \right) = \pm \sqrt{\frac{1 – \cos(A)}{1 + \cos(A)}}
\]

SL Loney – Examples 17

#4. Prove that \[
\frac{\sin(2A)}{1 + \cos(2A)} = \tan(A)
\]Solution: \[
LHS = \frac{2\sin(A)\cos(A)}{1 + 2\cos^2(A) – 1} = \frac{\sin(A)}{\cos(A)} = \tan(A) = RHS
\]

#5. Prove that \[
\frac{\sin(2A)}{1 – \cos(2A)} = \cot(A)
\]
#7. Prove that \[
\tan(A) + \cot(A) = 2\csc(2A)
\]Solution: \[
LHS = \frac{\sin(A)}{\cos(A)} + \frac{\cos(A)}{\sin(A)} = \frac{\sin^2(A) + \cos^2(A)}{\sin(A)\cos(A)}
\]\[
= \frac{1}{\sin(A)\cos(A)} = \frac{2}{2\sin(A)\cos(A)} = \frac{2}{\sin(2A)} = 2\csc(2A) = RHS
\]

#9. Prove that \[
\csc(2A) + \cot(2A) = \cot(A)
\]Solution: Here, LHS \[
= \frac{1}{\sin(2A)} + \frac{\cos(2A)}{\sin(2A)} = \frac{1 + \cos(2A)}{\sin(2A)}
\]\[
= \frac{1 + 2\cos^2(A) – 1}{2\sin(A)\cos(A)} = \frac{\cos(A)}{\sin(A)} = \cot(A) = RHS
\]

#15. To prove: \[
\frac{\sin^2(A) – \sin^2(B)}{\sin(A)\cos(A) – \sin(B)\cos(B)} = \tan(A + B)
\]Solution: LHS \(=\) \[
\frac{2\sin^2(A) – 2\sin^2(B)}{2\sin(A)\cos(A) – 2\sin(B)\cos(B)}
\]\[
= \frac{1 – \cos(2A) – 1 + \cos(2B)}{\sin(2A) – \sin(2B)} = \frac{\cos(2B) – \cos(2A)}{\sin(2A) – \sin(2B)} = \frac{2\sin(A + B)\sin(A – B)}{2\cos(A + B)\sin(A – B)}
\]\[
= \tan(A + B) = RHS
\]

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#17. Prove that \[
\frac{\cos(A) + \sin(A)}{\cos(A) – \sin(A)} – \frac{\cos(A) – \sin(A)}{\cos(A) + \sin(A)} = 2\tan(2A)
\] Solution: \[
LHS = \frac{(\cos A + \sin A)^2 – (\cos A – \sin A)^2}{cos^2 A – sin^2 A}
\]\[
= \frac{(cosA + sinA + cosA – sinA)(cosA + sinA – cosA + sinA)}{\cos(2A)}
\]\[
= \frac{2\cos(A) 2\sin(A)}{\cos(2A)} = \frac{2 \times 2\sin(A)\cos(A)}{\cos(2A)}
\]\[
= 2 \times \frac{\sin(2A)}{\cos(2A)} = 2\tan(2A) = RHS
\]

#24. Prove that \[
\frac{\sin(A + 3B) + \sin(3A + B)}{\sin(2A) + \sin(2B)} = 2\cos(A + B)
\]

#29. Prove: \[
\sec^2(A) (1 + \sec(2A)) = 2\sec(2A)
\]Solution: \[
LHS = \frac{1}{\cos^2(A)}\left( \frac{1 + \cos(2A)}{\cos(2A)} \right) = \frac{1 + 2\cos^2(A) – 1}{\cos^2(A)\cos(2A)}
\]\[
= \frac{2}{\cos(2A)} = 2\sec(2A) = RHS
\]

#35. To prove: \[
\cos(20{°})\cos(40{°})\cos(60{°})\cos(80{°}) = \frac{1}{16}
\]Solution: \[
LHS = \frac{1}{2}\cos(20{°})\cos(40{°})\cos(80{°})
\]\[
= \frac{1}{2} \times \frac{1}{2}\cos(20{°}) \times \{2\cos(80{°})\cos(40{°})\}
\]\[
= \frac{1}{4}\cos(20{°})\{\cos(120{°}) + \cos(40{°})\}
\]\[
= \frac{1}{4}\cos(20{°})\left\{ -\frac{1}{2} + \cos(40{°}) \right\}
\]\[
= \frac{1}{8}\cos(20{°})\{2\cos(40{°}) – 1\}
\]\[
= \frac{1}{8}[2\cos(40{°})\cos(20{°}) – \cos(20{°})]
\]\[
= \frac{1}{8}[\cos(60{°}) + \cos(20{°}) – \cos(20{°})]
\]\[
= \frac{1}{16} = RHS
\]

#36. To prove: \[
\sin(20{°})\sin(40{°})\sin(60{°})\sin(80{°}) = \frac{3}{16}
\]

#40. Prove: \[
\tan(3A)\tan(2A)\tan(A) = \tan(3A) – \tan(2A) – \tan(A)
\]

Solution: Starting with \(\tan3A\),

\[
\tan(3A) = \tan(2A + A) = \frac{\tan(2A) + \tan(A)}{1 – \tan(2A)\tan(A)}
\]\[
\Rightarrow \tan(3A) – \tan(3A)\tan(2A)\tan(A) = \tan(2A) + \tan(A)
\]\[
\Rightarrow \tan(3A) – \tan(2A) – \tan(A) = \tan(3A)\tan(2A)\tan(A)
\]

SL Loney – Examples 18

#12. Prove that \[
\sin(A) = \frac{2\tan\mkern-5mu\left( \frac{A}{2} \right)}{1 + \tan^2\mkern-5mu\left( \frac{A}{2} \right)}
\]Solution: \[
RHS = \frac{2\sin\mkern-5mu\left( \frac{A}{2} \right)}{\cos\mkern-5mu\left( \frac{A}{2} \right)} \div 1 + \frac{\sin^2\mkern-5mu\left( \frac{A}{2} \right)}{\cos^2\mkern-5mu\left( \frac{A}{2} \right)}
\]\[
= \frac{2\sin\mkern-5mu\left( \frac{A}{2} \right)}{\cos\mkern-5mu\left( \frac{A}{2} \right)} \div \frac{\cos^2\mkern-5mu\left( \frac{A}{2} \right) + \sin^2\mkern-5mu\left( \frac{A}{2} \right)}{\cos^2\mkern-5mu\left( \frac{A}{2} \right)}
\]\[
= 2\sin\mkern-5mu\left( \frac{A}{2} \right)\cos\mkern-5mu\left( \frac{A}{2} \right) = \sin(A) = LHS
\]

#13. Prove that \[
\cos(A) = \frac{1 – \tan\mkern-5mu\left( \frac{A}{2} \right)^2}{1 + \tan\mkern-5mu\left( \frac{A}{2} \right)^2}
\]
#16. Prove that \[
\sin^2\mkern-5mu\left( \frac{\pi}{8} + \frac{A}{2} \right) – \sin^2\mkern-5mu\left( \frac{\pi}{8} – \frac{A}{2} \right) = \frac{1}{\sqrt{2}}\sin(A)
\]Solution: \[
LHS = \frac{1}{2}\left[ 2\sin^2\mkern-5mu\left( \frac{\pi}{8} + \frac{A}{2} \right) – 2\sin^2\mkern-5mu\left( \frac{\pi}{8} – \frac{A}{2} \right) \right]
\]Since \(1 – \cos(2A) = 2\sin^2(A),\)\[
LHS = \frac{1}{2}\left[ 1 – \cos\mkern-5mu\left( \frac{\pi}{4} + A \right) – \left\{ 1 – \cos\mkern-5mu\left( \frac{\pi}{4} – A \right) \right\} \right]
\]\[
= \frac{1}{2}\left[ \cos\mkern-5mu\left( \frac{\pi}{4} – A \right) – \cos\mkern-5mu\left( \frac{\pi}{4} + A \right) \right]
\]\[
= \frac{1}{2} \times 2\sin\mkern-5mu\left( \frac{\pi}{4} \right)\sin(A) = \frac{1}{\sqrt{2}}\sin(A)
\]

SL Loney – Examples 19

#2. Find the value of \[
\cos^2(48{°}) – \sin^2(12{°})
\]Solution: Given expression \[
= \cos(48{°} + 12{°}) + \cos(48{°} – 12{°})
\]\[
= \cos(60{°})\cos(36{°})
\]\[
= \frac{1}{2} \times \frac{\sqrt{5 + 1}}{4} = \frac{\sqrt{5 + 1}}{8}
\]

#4. Find the value of \[
\sin\mkern-5mu\left( \frac{\pi}{5} \right)\sin\mkern-5mu\left( \frac{2\pi}{5} \right)\sin\mkern-5mu\left( \frac{3\pi}{5} \right)\sin\mkern-5mu\left( \frac{4\pi}{5} \right)
\]Solution: Given expression \[
= \sin(36{°})\sin(72{°})\sin(108{°})\sin(144{°})
\]\[
= \sin(36{°})\sin(72{°})\sin(72{°})\sin(36{°})
\]\[
= \sin^2(36{°})\sin^2(72{°})
\]\[
= \frac{5 – \sqrt{5}}{8} \times \frac{5 + \sqrt{5}}{8} = \frac{25 – 5}{64} = \frac{5}{16}
\]

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