14. -/1 points details my notes sprecalc7 6.2.032. evaluate the expression without using a calculator. (sin…
14. -/1 points details my notes sprecalc7 6.2.032. evaluate the expression without using a calculator. (sin 60°)² + (cos 60°)² 15. -/1 points details my notes sprecalc7 6.2.034.mi. evaluate the expression without using a calculator. (sin(π/3)cos(π/4) - sin(π/4)cos(π/3))² 16. -/1 points details my notes sprecalc7 6.2.036. evaluate the expression without using a calculator. (sin(π/3)tan(π/6)+csc(π/4))²
Answer
Explanation:
Step1: Recall trigonometric values
We know that $\sin60^{\circ}=\frac{\sqrt{3}}{2}$ and $\cos60^{\circ}=\frac{1}{2}$.
Step2: Substitute values into the expression
$(\sin60^{\circ})^2+(\cos60^{\circ})^2 = (\frac{\sqrt{3}}{2})^2+(\frac{1}{2})^2$.
Step3: Calculate the squares
$(\frac{\sqrt{3}}{2})^2=\frac{3}{4}$ and $(\frac{1}{2})^2=\frac{1}{4}$.
Step4: Add the results
$\frac{3}{4}+\frac{1}{4}=\frac{3 + 1}{4}=1$.
Answer:
1
Explanation:
Step1: Recall trigonometric values
$\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{3}=\frac{1}{2}$.
Step2: Substitute values into the expression
$(\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\sin\frac{\pi}{4}\cos\frac{\pi}{3})^2=(\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2})^2$.
Step3: Simplify the product - terms
$\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}}{4}$ and $\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{2}}{4}$.
Step4: Subtract the terms inside the parentheses
$\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}$.
Step5: Square the result
$(\frac{\sqrt{6}-\sqrt{2}}{4})^2=\frac{(\sqrt{6}-\sqrt{2})^2}{16}=\frac{6 - 2\sqrt{12}+2}{16}=\frac{8 - 4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}$.
Answer:
$\frac{2 - \sqrt{3}}{4}$
Explanation:
Step1: Recall trigonometric values
$\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$, $\csc\frac{\pi}{4}=\sqrt{2}$.
Step2: Substitute values into the expression
$(\sin\frac{\pi}{3}\tan\frac{\pi}{6}+\csc\frac{\pi}{4})^2=(\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{3}+\sqrt{2})^2$.
Step3: Simplify the product - term
$\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{3}=\frac{3}{6}=\frac{1}{2}$.
Step4: Add the terms inside the parentheses
$\frac{1}{2}+\sqrt{2}=\frac{1 + 2\sqrt{2}}{2}$.
Step5: Square the result
$(\frac{1 + 2\sqrt{2}}{2})^2=\frac{1+4\sqrt{2}+8}{4}=\frac{9 + 4\sqrt{2}}{4}$.
Answer:
$\frac{9 + 4\sqrt{2}}{4}$