cot θ = 12 / 5 = > tan θ = 1 / cot θ = 5 / 12 \cot\theta=12/5=>\tan \theta=1/\cot \theta=5/12 cot θ = 12/5 => tan θ = 1/ cot θ = 5/12 1.
If cos θ = 1 / tan θ , \cos θ= 1/\tan θ, cos θ = 1/ tan θ , cos θ = 1 / tan θ = 12 / 5. \cos θ= 1/\tan θ=12/5. cos θ = 1/ tan θ = 12/5.
There is no solution, because − 1 ≤ cos θ ≤ 1 -1\leq \cos \theta\leq 1 − 1 ≤ cos θ ≤ 1 θ ∈ R . \theta \in \R. θ ∈ R .
2.
Suppose that the condition cos θ = 1 / tan θ \cos θ= 1/\tan θ cos θ = 1/ tan θ
Then
1 + cot 2 θ = 1 sin 2 θ 1+\cot^2\theta=\dfrac{1}{\sin^2\theta} 1 + cot 2 θ = sin 2 θ 1
sin 2 θ = 1 1 + cot 2 θ = 1 1 + ( 12 / 5 ) 2 = 25 169 \sin ^2 \theta=\dfrac{1}{1+\cot^2\theta}=\dfrac{1}{1+(12/5)^2}=\dfrac{25}{169} sin 2 θ = 1 + cot 2 θ 1 = 1 + ( 12/5 ) 2 1 = 169 25
cos 2 θ = 1 − sin 2 θ = 1 − 25 169 = 144 169 \cos ^2\theta=1-\sin^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169} cos 2 θ = 1 − sin 2 θ = 1 − 169 25 = 169 144
cos ( 2 θ ) cos θ + sin θ = cos 2 θ − sin 2 θ cos θ + sin θ \dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\dfrac{\cos^2 \theta-\sin^2 \theta}{\cos \theta+\sin \theta} cos θ + sin θ cos ( 2 θ ) = cos θ + sin θ cos 2 θ − sin 2 θ
= ( cos θ − sin θ ) ( cos θ + sin θ ) cos θ + sin θ =\dfrac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta+\sin \theta} = cos θ + sin θ ( cos θ − sin θ ) ( cos θ + sin θ )
= cos θ − sin θ , cos θ ≠ − sin θ =\cos \theta-\sin \theta,\cos \theta\not=-\sin \theta = cos θ − sin θ , cos θ = − sin θ i)
sin θ > 0 , cos θ > 0 \sin\theta>0, \cos \theta>0 sin θ > 0 , cos θ > 0 Then
sin θ = 25 169 = 5 13 \sin \theta=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13} sin θ = 169 25 = 13 5
cos θ = 144 169 = 12 13 \cos \theta=\sqrt{\dfrac{144}{169}}=\dfrac{12}{13} cos θ = 169 144 = 13 12
cos ( 2 θ ) cos θ + sin θ = cos θ − sin θ \dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\cos \theta-\sin \theta cos θ + sin θ cos ( 2 θ ) = cos θ − sin θ
= 12 13 − 5 13 = 8 13 =\dfrac{12}{13}-\dfrac{5}{13}=\dfrac{8}{13} = 13 12 − 13 5 = 13 8
ii)
sin θ < 0 , cos θ < 0 \sin\theta<0, \cos \theta<0 sin θ < 0 , cos θ < 0 Then
sin θ = − 25 169 = − 5 13 \sin \theta=-\sqrt{\dfrac{25}{169}}=-\dfrac{5}{13} sin θ = − 169 25 = − 13 5
cos θ = − 144 169 = − 12 13 \cos \theta=-\sqrt{\dfrac{144}{169}}=-\dfrac{12}{13} cos θ = − 169 144 = − 13 12
cos ( 2 θ ) cos θ + sin θ = cos θ − sin θ \dfrac{\cos(2\theta)}{\cos \theta+\sin \theta}=\cos \theta-\sin \theta cos θ + sin θ cos ( 2 θ ) = cos θ − sin θ
= − 12 13 − ( − 5 13 ) = − 8 13 =-\dfrac{12}{13}-(-\dfrac{5}{13})=-\dfrac{8}{13} = − 13 12 − ( − 13 5 ) = − 13 8
#340153 on Dec 2023
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