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Question Video: Using Sum and Difference of Angles Identities to Evaluate Trigonometric Expressions Involving Special Angles Mathematics • 10th Grade

Evaluate sin (πœ‹/3) cos (πœ‹/3) + cos (πœ‹/3) sin (πœ‹/3).

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Video Transcript

Evaluate sin of πœ‹ over three times cos of πœ‹ over three plus cos of πœ‹ over three times sin of πœ‹ over three.

To answer this question, we will first need to evaluate sin of πœ‹ over three and cos of πœ‹ over three. Then, we will substitute those values into the original expression. It is worth noting that the original expression can be simplified as two times sin of πœ‹ over three times cos of πœ‹ over three. We recall that πœ‹ over three equals 60 degrees. We also know that 60 degrees in standard position terminates in the first quadrant, where both the π‘₯- and 𝑦-coordinates are positive. Therefore, all trigonometric functions in the first quadrant are positive, including sin and cos of 60 degrees.

We will now use our knowledge of 30-60-90 special right triangles to evaluate sin of 60 degrees and cos of 60 degrees. We recall that the hypotenuse of a 30-60-90 right triangle is twice the length of the side opposite the 30-degree angle. We label these sides 2π‘Ž and π‘Ž. The side opposite the 60-degree angle has a length equal to the square root of three times the length of the side opposite the 30-degree angle. So we label that side as π‘Ž times the square root of three. We now sketch the 30-60-90 triangle as a reference triangle on our coordinate plane, where the hypotenuse is the terminal side of our 60-degree angle in standard position.

We’ve recalled the coordinate definition of sine is 𝑦 over π‘Ÿ and cosine is π‘₯ over π‘Ÿ, where π‘Ÿ is the distance from the origin to the point π‘₯, 𝑦. π‘Ÿ also represents the length of the hypotenuse of the reference triangle. If we use an π‘Ÿ-value of one, then the definition of sine reduces to the 𝑦-value and the definition of cosine reduces to the π‘₯-value. To solve for π‘₯ and 𝑦, we go back to the established relationships between the side lengths of any 30-60-90 triangle. In this case, the hypotenuse 2π‘Ž equals one. Therefore, by dividing each side of the equation by two, we find that π‘Ž equals one-half. Now that we have the value of π‘Ž, we can solve for π‘₯ and 𝑦.

By substituting the value of π‘Ž, we find that π‘₯ equals one-half and 𝑦 equals one-half times square root of three, which simplifies to the square root of three over two. So, sin of 60 degrees equals square root of three over two and cos of 60 degrees equals one-half. As expected, both sin and cos of 60 degrees, or πœ‹ over three, have positive values. To finish answering the question, we substitute the positive values we found for sine and cosine into the original expression. The result is square root of three over two times one-half plus one-half times the square root of three over two or two times the product of square root of three over two and one-half.

Following order of operations, we multiply first. And square root of three over four plus the square root of three over four equals two square root of three over four, which simplifies to our final answer of square root of three over two. In conclusion, the value of sin of πœ‹ over three times cos of πœ‹ over three plus cos of πœ‹ over three times sin of πœ‹ over three is the square root of three over two.

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