Question

Find the exact value of each expression. Do not use a calculator. $$ \sin ^{2} 60^{\circ}+\cos ^{2} 30^{\circ} $$

Short answer

\( \frac{3}{2} \)

Step-by-step solution

Identify known trigonometric values

Recall the exact values of \( \sin 60^{\circ} \) and \( \cos 30^{\circ} \). We know from trigonometric tables that \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) and \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \).

Square each trigonometric value

Next, square the values obtained: \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \Rightarrow (\frac{\sqrt{3}}{2})^2 \) and \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \Rightarrow (\frac{\sqrt{3}}{2})^2 \).

Compute the squared values

Calculate the squares: \( (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} \). Therefore: \( \sin^{2} 60^{\circ} = \frac{3}{4} \) and \( \cos^{2} 30^{\circ} = \frac{3}{4} \).

Add the squared values

Finally, add the squared values: \( \sin^{2} 60^{\circ} + \cos^{2} 30^{\circ} = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2} \).

Key concepts

sine function

The sine function, often abbreviated as \(\text{sin} \), is a fundamental trigonometric function. It relates the angle of a right triangle to the ratio of the length of the opposite side to the hypotenuse. Mathematically, for an angle \( \theta \), we write:
\( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \).
This function is periodic with a period of \( 360^{\circ} \) or \( 2\pi \) radians. Important exact values to remember include:

• \( \sin 0^{\circ} = 0 \)
• \( \sin 30^{\circ} = \frac{1}{2} \)
• \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)
• \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
• \( \sin 90^{\circ} = 1 \)

These exact values are useful for problems requiring non-calculator solutions.

cosine function

The cosine function, abbreviated as \( \cos \), is another primary trigonometric function. It is closely related to the sine function and measures the ratio of the length of the adjacent side to the hypotenuse in a right triangle. Mathematically, for an angle \( \theta \):
\( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \).
Like the sine function, the cosine function is also periodic with a period of \( 360^{\circ} \) or \( 2\pi \) radians. Essential exact values include:

• \( \cos 0^{\circ} = 1 \)
• \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
• \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \)
• \( \cos 60^{\circ} = \frac{1}{2} \)
• \( \cos 90^{\circ} = 0 \)

Knowing these values helps in various algebraic and trigonometric manipulations.

exact trigonometric values

Exact trigonometric values are critical in solving problems without a calculator. They represent specific angles where the sine and cosine functions produce simple rational outputs. Typically, these values are derived from special triangles like the 30-60-90 and 45-45-90 triangles. For instance:

• For a 45-45-90 triangle, sides are in the ratio \( 1:1:\sqrt{2} \) leading to:
  • \( \sin 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \)
  • \( \cos 45^{\circ} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \)
• For a 30-60-90 triangle, sides are in the ratio \( 1:\sqrt{3}:2 \) leading to:
  • \( \sin 30^{\circ} = \frac{1}{2} \)
  • \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
  • \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)
  • \( \cos 60^{\circ} = \frac{1}{2} \)

These values simplify exploring more complex trigonometric identities and equations.

algebraic manipulation

Algebraic manipulation involves rearranging and simplifying algebraic expressions. It is an essential skill in mathematics, particularly in solving trigonometric problems. Here is how we use algebraic manipulation in our problem:

• First, recall the values: \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) and \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
• Next, we square each value:
  \( \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \)
  This makes both \( \sin^{2} 60^{\circ} \) and \( \cos^{2} 30^{\circ} \) equal to \( \frac{3}{4} \)
• Finally, add the squared values:
  \( \sin^{2} 60^{\circ} + \cos^{2} 30^{\circ} = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2} \)

This careful step-by-step algebraic manipulation ensures accurate results.