If sin α = 1 2 \sin\alpha=\frac{1}{2} sin α = 2 1 and 0 < α < π 2 , 0<\alpha<\frac{\pi}{2}, 0 < α < 2 π , what is the value of cos 2 α ? \cos2\alpha? cos 2 α ?
Given sin α = 1 2 \sin\alpha=\frac{1}{2} sin α = 2 1 and 0 < α < π 2 , 0<\alpha<\frac{\pi}{2}, 0 < α < 2 π , we have cos α = 1 − sin 2 α = 1 − ( 1 2 ) 2 = 3 2 . \cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\left(\frac{1}{2}\right)^2}=\frac{\sqrt{3}}{2}. cos α = 1 − sin 2 α = 1 − ( 2 1 ) 2 = 2 3 . This implies cos 2 α = cos 2 α − sin 2 α = ( 3 2 ) 2 − ( 1 2 ) 2 = 3 4 − 1 4 = 1 2 . □ \cos2\alpha=\cos^2\alpha-\sin^2\alpha=\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}. \ _\square cos 2 α = cos 2 α − sin 2 α = ( 2 3 ) 2 − ( 2 1 ) 2 = 4 3 − 4 1 = 2 1 . □
Use the double angle identity to solve the following problems:
Submit your answer
Suppose
cos θ = 3 5 , sin 2 θ = a b , \cos\theta = \frac{3}{5},\quad \sin2\theta = \frac{a}{b}, cos θ = 5 3 , sin 2 θ = b a ,
where 0 < θ < π 2 0<\theta <\frac{\pi}{2} 0 < θ < 2 π with a a a and b b b coprime positive integers.
What is the value of a + b ? a+b? a + b ?
If sin θ = 5 13 \sin\theta=\frac{5}{13} sin θ = 1 3 5 and 0 < θ < π 2 , 0<\theta<\frac{\pi}{2}, 0 < θ < 2 π , what is the value of sin 2 θ ? \sin2\theta? sin 2 θ ?
Given sin θ = 5 13 \sin\theta=\frac{5}{13} sin θ = 1 3 5 and 0 < θ < π 2 , 0<\theta<\frac{\pi}{2}, 0 < θ < 2 π , we have cos θ = 1 − sin 2 θ = 1 − ( 5 13 ) 2 = 12 13 . \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\frac{5}{13}\right)^2}=\frac{12}{13}. cos θ = 1 − sin 2 θ = 1 − ( 1 3 5 ) 2 = 1 3 1 2 . This implies sin 2 θ = 2 sin θ cos θ = 2 ⋅ 5 13 ⋅ 12 13 = 120 169 . □ \sin2\theta=2\sin\theta\cos\theta =2\cdot \frac{5}{13} \cdot \frac{12}{13}=\frac{120}{169}. \ _\square sin 2 θ = 2 sin θ cos θ = 2 ⋅ 1 3 5 ⋅ 1 3 1 2 = 1 6 9 1 2 0 . □
If sin α = 3 5 \sin\alpha=\frac{3}{5} sin α = 5 3 and π 2 < α < π , \frac{\pi}{2}<\alpha<\pi, 2 π < α < π , what is the value of tan 2 α ? \tan2\alpha? tan 2 α ?
Given sin α = 3 5 \sin\alpha=\frac{3}{5} sin α = 5 3 and π 2 < α < π , \frac{\pi}{2}<\alpha<\pi, 2 π < α < π , we have cos α = − 1 − sin 2 α = − 1 − ( 3 5 ) 2 = − 4 5 . \cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-\left(\frac{3}{5}\right)^2}=-\frac{4}{5}. cos α = − 1 − sin 2 α = − 1 − ( 5 3 ) 2 = − 5 4 . This implies tan α = sin α cos α = − 3 4 . \displaystyle{\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=-\frac{3}{4}}. tan α = cos α sin α = − 4 3 . Hence, it follows that tan 2 α = 2 tan α 1 − tan 2 α = 2 ⋅ ( − 3 4 ) 1 − ( − 3 4 ) 2 = − 24 7 . □ \tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\cdot\left(-\frac{3}{4}\right)}{1-\left(-\frac{3}{4}\right)^2}=-\frac{24}{7}. \ _\square tan 2 α = 1 − tan 2 α 2 tan α = 1 − ( − 4 3 ) 2 2 ⋅ ( − 4 3 ) = − 7 2 4 . □
What is θ \theta θ ( 0 < θ < π ) (0<\theta<\pi) ( 0 < θ < π ) that satisfies the following:
tan 2 θ + sec 2 θ − 1 tan 2 θ + sec 2 θ + 1 = − 1 ? \frac{\tan2\theta+\sec2\theta-1}{\tan2\theta+\sec2\theta+1}=-1? tan 2 θ + sec 2 θ + 1 tan 2 θ + sec 2 θ − 1 = − 1 ?
Observe that the given equation can be rewritten as follows:
tan 2 θ + sec 2 θ − 1 tan 2 θ + sec 2 θ + 1 = sin 2 θ cos 2 θ + 1 cos 2 θ − 1 sin 2 θ cos 2 θ + 1 cos 2 θ + 1 = sin 2 θ + 1 − cos 2 θ sin 2 θ + 1 + cos 2 θ = 2 sin θ cos θ + 2 sin 2 θ 2 sin θ cos θ + 2 cos 2 θ = sin θ ( cos θ + sin θ ) cos θ ( sin θ + cos θ ) = tan θ = − 1. \begin{aligned}
\frac{\tan2\theta+\sec2\theta-1}{\tan2\theta+\sec2\theta+1}
&=\frac{\frac{\sin2\theta}{\cos2\theta}+\frac{1}{\cos2\theta}-1}{\frac{\sin2\theta}{\cos2\theta}+\frac{1}{\cos2\theta}+1}\\\\
&=\frac{\sin2\theta+1-\cos2\theta}{\sin2\theta+1+\cos2\theta}\\\\
&=\frac{2\sin\theta\cos\theta+2\sin^2\theta}{2\sin\theta\cos\theta+2\cos^2\theta}\\\\
&=\frac{\sin\theta(\cos\theta+\sin\theta)}{\cos\theta(\sin\theta+\cos\theta)}\\\\
&=\tan\theta=-1.
\end{aligned} tan 2 θ + sec 2 θ + 1 tan 2 θ + sec 2 θ − 1 = c o s 2 θ s i n 2 θ + c o s 2 θ 1 + 1 c o s 2 θ s i n 2 θ + c o s 2 θ 1 − 1 = sin 2 θ + 1 + cos 2 θ sin 2 θ + 1 − cos 2 θ = 2 sin θ cos θ + 2 cos 2 θ 2 sin θ cos θ + 2 sin 2 θ = cos θ ( sin θ + cos θ ) sin θ ( cos θ + sin θ ) = tan θ = − 1 .
Then, since 0 < θ < π , 0<\theta<\pi, 0 < θ < π , θ = 3 π 4 . \theta=\frac{3\pi}{4}. θ = 4 3 π . □ _\square □
If csc α = 2 3 3 \csc\alpha=\frac{2\sqrt{3}}{3} csc α = 3 2 3 for 0 < α < π 2 , 0<\alpha<\frac{\pi}{2}, 0 < α < 2 π , what is sin 2 α + cos 2 α ? \sin2\alpha+\cos2\alpha? sin 2 α + cos 2 α ?
Since sin α = 1 csc α = 3 2 3 = 3 2 \sin\alpha=\frac{1}{\csc\alpha}=\frac{3}{2\sqrt{3}}=\frac{\sqrt{3}}{2} sin α = c s c α 1 = 2 3 3 = 2 3 and 0 < α < π 2 , 0<\alpha<\frac{\pi}{2}, 0 < α < 2 π , α = π 3 \alpha=\frac{\pi}{3} α = 3 π and cos α = 1 2 . \cos\alpha=\frac{1}{2}. cos α = 2 1 . Hence,
sin 2 α + cos 2 α = 2 sin α cos α + ( cos 2 α − sin 2 α ) = 2 ⋅ 3 2 ⋅ 1 2 + ( 1 2 ) 2 − ( 3 2 ) 2 = 3 2 + 1 4 − 3 4 = 3 − 1 2 . □ \begin{aligned}
\sin2\alpha+\cos2\alpha
&=2\sin\alpha\cos\alpha+\big( \cos^2\alpha - \sin^2\alpha \big) \\
&=2\cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2}+ \left(\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2 \\
&=\frac{\sqrt{3}}{2}+\frac{1}{4}-\frac{3}{4} \\
&=\frac{\sqrt{3}-1}{2}. \ _\square
\end{aligned} sin 2 α + cos 2 α = 2 sin α cos α + ( cos 2 α − sin 2 α ) = 2 ⋅ 2 3 ⋅ 2 1 + ( 2 1 ) 2 − ( 2 3 ) 2 = 2 3 + 4 1 − 4 3 = 2 3 − 1 . □
What is X X X in the following identity: 1 + cot 2 θ = X sin 2 2 θ ? \displaystyle{1+\cot^2\theta=\frac{X}{\sin^2 2\theta}}? 1 + cot 2 θ = sin 2 2 θ X ?
( a ) cos 2 θ ( b ) 4 cos 2 θ ( c ) sin 2 θ ( d ) 4 sin 2 θ \begin{array}{c}(a) \cos^2\theta &&&(b)\ 4\cos^2\theta &&&(c) \sin^2\theta &&&(d)\ 4\sin^2\theta \end{array} ( a ) cos 2 θ ( b ) 4 cos 2 θ ( c ) sin 2 θ ( d ) 4 sin 2 θ
Since 1 + cot 2 θ = csc 2 θ 1+\cot^2\theta=\csc^2\theta 1 + cot 2 θ = csc 2 θ , we have
csc 2 θ = X sin 2 2 θ ( sin 2 θ sin θ ) 2 = X . \begin{aligned}
\csc^2\theta&=\frac{X}{\sin^22\theta}\\
\left(\frac{\sin2\theta}{\sin\theta}\right)^2&=X.
\end{aligned} csc 2 θ ( sin θ sin 2 θ ) 2 = sin 2 2 θ X = X .
Using the identity sin 2 θ = 2 sin θ cos θ \sin2\theta=2\sin\theta\cos\theta sin 2 θ = 2 sin θ cos θ gives
X = ( sin 2 θ sin θ ) 2 = ( 2 sin θ cos θ sin θ ) 2 = 4 cos 2 θ . X=\left(\frac{\sin2\theta}{\sin\theta}\right)^2=\left(\frac{2\sin\theta\cos\theta}{\sin\theta}\right)^2=4\cos^2\theta. X = ( sin θ sin 2 θ ) 2 = ( sin θ 2 sin θ cos θ ) 2 = 4 cos 2 θ .
Therefore, the answer is ( b ) . (b). ( b ) . □ _\square □
Use the double angle formulas to prove the identity
csc 2 θ − cot 2 θ = tan θ . \csc 2\theta - \cot 2\theta = \tan \theta. csc 2 θ − cot 2 θ = tan θ .
We have
csc 2 θ − cot 2 θ = 1 sin 2 θ − cos 2 θ sin 2 θ = 1 − cos 2 θ sin 2 θ . \begin{aligned}
\csc 2\theta - \cot 2\theta
&= \dfrac{1}{\sin 2\theta} - \dfrac{\cos 2\theta}{\sin 2\theta} \\ \\
&= \dfrac{1-\cos 2\theta}{\sin 2\theta}.
\end{aligned} csc 2 θ − cot 2 θ = sin 2 θ 1 − sin 2 θ cos 2 θ = sin 2 θ 1 − cos 2 θ .
Using the identities 1 − cos 2 θ = 2 sin 2 θ 1-\cos 2\theta = 2 \sin^2 \theta 1 − cos 2 θ = 2 sin 2 θ and sin 2 θ = 2 sin θ cos θ \sin 2\theta = 2\sin \theta \cos \theta sin 2 θ = 2 sin θ cos θ gives
csc 2 θ − cot 2 θ = 1 − cos 2 θ sin 2 θ = 2 sin 2 θ 2 sin θ cos θ = sin θ cos θ = tan θ . □ \csc 2\theta - \cot 2\theta = \dfrac{1-\cos 2\theta}{\sin 2\theta} = \dfrac{2 \sin^2 \theta}{2 \sin \theta \cos \theta} = \dfrac{\sin \theta}{\cos \theta} = \tan \theta. \ _\square csc 2 θ − cot 2 θ = sin 2 θ 1 − cos 2 θ = 2 sin θ cos θ 2 sin 2 θ = cos θ sin θ = tan θ . □