Conditions
1) If tan x = . 76 \tan x = .76 tan x = .76 and cos x = . 22 \cos x = .22 cos x = .22 , what is sin x \sin x sin x ?
2) If sin x = . 42 \sin x = .42 sin x = .42 , what is cos x \cos x cos x ?
Solution
1)
This is a trigonometric task. Here we must use the Basic Trigonometric Identity. Its formula is:
sin 2 ( x ) + cos 2 ( x ) = 1 , x ∈ R . \sin^2(x) + \cos^2(x) = 1, \, x \in \mathbb{R}. sin 2 ( x ) + cos 2 ( x ) = 1 , x ∈ R .
As it known, tan ( x ) = sin ( x ) cos ( x ) \tan(x) = \frac{\sin(x)}{\cos(x)} tan ( x ) = c o s ( x ) s i n ( x ) , so
sin ( x ) cos ( x ) = 0.76 ⇒ sin ( x ) = 0.76 cos ( x ) . cos ( x ) = 0.22 ⇒ sin ( x ) = 0.76 × 0.22 = 0.1672 \frac{\sin(x)}{\cos(x)} = 0.76 \Rightarrow \sin(x) = 0.76 \cos(x). \quad \cos(x) = 0.22 \Rightarrow \sin(x) = 0.76 \times 0.22 = 0.1672 cos ( x ) sin ( x ) = 0.76 ⇒ sin ( x ) = 0.76 cos ( x ) . cos ( x ) = 0.22 ⇒ sin ( x ) = 0.76 × 0.22 = 0.1672
2)
This task is logically near previous. We use the same formula:
sin 2 ( x ) + cos 2 ( x ) = 1 , x ∈ R . \sin^2(x) + \cos^2(x) = 1, \, x \in \mathbb{R}. sin 2 ( x ) + cos 2 ( x ) = 1 , x ∈ R . sin 2 ( x ) + cos 2 ( x ) = ( 0.42 ) 2 + cos 2 ( x ) = 1 ⇒ cos 2 ( x ) = 1 − 0.1764 = 0.8236 ⇒ cos ( x ) = ± 0.8236 \begin{array}{l}
\sin^2(x) + \cos^2(x) = (0.42)^2 + \cos^2(x) = 1 \Rightarrow \cos^2(x) = 1 - 0.1764 = 0.8236 \\
\Rightarrow \cos(x) = \pm \sqrt{0.8236} \\
\end{array} sin 2 ( x ) + cos 2 ( x ) = ( 0.42 ) 2 + cos 2 ( x ) = 1 ⇒ cos 2 ( x ) = 1 − 0.1764 = 0.8236 ⇒ cos ( x ) = ± 0.8236
This value is irrational, so we can leave it in this form.
Answer:
1) sin ( x ) = 0.1672 \sin(x) = 0.1672 sin ( x ) = 0.1672
2) Two solutions: cos ( x ) = 0.8236 \cos(x) = \sqrt{0.8236} cos ( x ) = 0.8236 or cos ( x ) = − 0.8236 \cos(x) = -\sqrt{0.8236} cos ( x ) = − 0.8236