35. 2/2 points details write the trigonometric expression as an algebraic expression. cos(arccos(2x)…
35. 2/2 points details write the trigonometric expression as an algebraic expression. cos(arccos(2x) - arcsin(2x)) 4x√1 - 4x² need help? read it submit answer 36. -/2 points details my notes larpcalclimaga8 5.4.054. 0/6 subm write the trigonometric expression as an algebraic expression. cos(arcsin(7x) - arctan(5x)) need help? read it submit answer 37. 2/2 points details my notes larpcalclimaga8 5.4.065. 1/6 subm write the expression as a trigonometric function of only θ, and use a graphing utility to confirm yo cos(7π/2 - θ) -sin(θ)
Answer
Explanation:
Step1: Recall the cosine - difference formula
$\cos(A - B)=\cos A\cos B+\sin A\sin B$. Let $A = \arcsin(7x)$ and $B=\arctan(5x)$.
Step2: Find $\cos A$ and $\sin A$ for $A=\arcsin(7x)$
If $A = \arcsin(7x)$, then $\sin A=7x$. Using the identity $\sin^{2}A+\cos^{2}A = 1$, we get $\cos A=\sqrt{1 - (7x)^{2}}=\sqrt{1 - 49x^{2}}$ (since the range of $\arcsin$ function gives non - negative cosine values in the principal range).
Step3: Find $\cos B$ and $\sin B$ for $B = \arctan(5x)$
If $B=\arctan(5x)=\frac{\sin B}{\cos B}$, and $\sin^{2}B+\cos^{2}B = 1$. Let $\tan B = 5x=\frac{y}{x}$ (in a right - triangle context), then $y = 5x$ and $r=\sqrt{1 + 25x^{2}}$. So $\sin B=\frac{5x}{\sqrt{1 + 25x^{2}}}$ and $\cos B=\frac{1}{\sqrt{1+25x^{2}}}$.
Step4: Substitute into the cosine - difference formula
$\cos(A - B)=\cos A\cos B+\sin A\sin B=\sqrt{1 - 49x^{2}}\cdot\frac{1}{\sqrt{1 + 25x^{2}}}+7x\cdot\frac{5x}{\sqrt{1 + 25x^{2}}}=\frac{\sqrt{1 - 49x^{2}}+35x^{2}}{\sqrt{1 + 25x^{2}}}$
Answer:
$\frac{\sqrt{1 - 49x^{2}}+35x^{2}}{\sqrt{1 + 25x^{2}}}$