give the exact value of the expression without using a calculator.\\n\\n\\( \\cos \\left( 2 \\arctan \\left(…

give the exact value of the expression without using a calculator.\\n\\n\\( \\cos \\left( 2 \\arctan \\left( - \\frac { 3 } { 4 } \\right) \\right) \\)\\n\\n\\( \\cos \\left( 2 \\arctan \\left( - \\frac { 3 } { 4 } \\right) \\right) = \\square \\)\\n(simplify your answer, including any radicals. use integers or fractions for any

give the exact value of the expression without using a calculator.\\n\\n\\( \\cos \\left( 2 \\arctan \\left( - \\frac { 3 } { 4 } \\right) \\right) \\)\\n\\n\\( \\cos \\left( 2 \\arctan \\left( - \\frac { 3 } { 4 } \\right) \\right) = \\square \\)\\n(simplify your answer, including any radicals. use integers or fractions for any

Answer

Explanation:

Step1: Let (\theta=\arctan\left(-\frac{3}{4}\right))

By the definition of the arctangent function, (\tan\theta =-\frac{3}{4}). We can consider a right - triangle where the opposite side (y=- 3) and the adjacent side (x = 4). Then, by the Pythagorean theorem (r=\sqrt{x^{2}+y^{2}}=\sqrt{4^{2}+\left(-3\right)^{2}}=\sqrt{16 + 9}=5). So, (\cos\theta=\frac{4}{5}) and (\sin\theta=-\frac{3}{5}).

Step2: Use the double - angle formula (\cos(2\alpha)=\cos^{2}\alpha-\sin^{2}\alpha)

Since (\alpha = \arctan\left(-\frac{3}{4}\right)), and we know (\cos\alpha=\frac{4}{5}), (\sin\alpha=-\frac{3}{5}). Then (\cos\left(2\arctan\left(-\frac{3}{4}\right)\right)=\cos^{2}\theta-\sin^{2}\theta). Substitute (\cos\theta=\frac{4}{5}) and (\sin\theta =-\frac{3}{5}) into the formula: [ \begin{align*} \cos^{2}\theta-\sin^{2}\theta&=\left(\frac{4}{5}\right)^{2}-\left(-\frac{3}{5}\right)^{2}\ &=\frac{16}{25}-\frac{9}{25}\ &=\frac{16 - 9}{25}\ &=\frac{7}{25} \end{align*} ]

Answer:

(\frac{7}{25})