Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions | Nagwa Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions | Nagwa

# Question Video: Using Trigonometric Values of Special Angles to Evaluate Trigonometric Expressions Mathematics • First Year of Secondary School

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Which of the following values of π₯ does not satisfy the equation sinΒ² (π₯) + 6 sin (3π₯) β 0.5 = 3β2? [A] β315Β° (B) 45Β° (C) 135Β° (D) 315Β° or (E) 405Β°

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### Video Transcript

Which of the following values of π₯ does not satisfy the equation sin squared π₯ plus six sin three π₯ minus 0.5 is equal to three root two? Is it (A) negative 315 degrees, (B) 45 degrees, (C) 135 degrees, (D) 315 degrees, or (E) 405 degrees?

In order to answer this question, we will use our knowledge of the special angles together with the properties of related angles when dealing with the sine function. We will begin by considering the five angles we are given and their corresponding positions on the CAST diagram. As positive angles are measured in the counterclockwise direction from the positive π₯-axis, 45 degrees lies in the first quadrant as shown. Adding 360 to 45 gives us 405. And this means that 405 degrees has the same position on the diagram. Likewise, subtracting 360 from 45 gives us negative 315. So, negative 315 degrees also lies in this position.

Following on from this, we know that sin of 45 degrees is equal to sin of 405 degrees, which is equal to sin of negative 315 degrees. Recalling that 45 degrees is one of our special angles and the sin of 45 degrees is equal to root two over two, then the sin of 405 degrees and the sin of negative 315 degrees must also equal root two over two. The two other angles we are given, 135 degrees and 315 degrees, lie in the second and fourth quadrants, respectively. In the second quadrant, the sine of any angle is positive, whereas in the fourth quadrant, the sine of any angle is negative.

Using the symmetry of the sine function and the fact that sin of 180 degrees minus π is equal to sin π and sin of 360 degrees minus π is equal to negative sin π, we see that sin of 135 degrees is also equal to root two over two. And sin of 315 degrees is equal to negative root two over two.

We are now nearly in a position to substitute our values into the equation. Before doing so, we notice that our second term includes sin of three π₯. This means that we need to multiply the angles by three before calculating their sine. Three multiplied by 45 is 135. And we already know that the sin of 135 degrees is root two over two. In the same way, three multiplied by 135 is 405. And the sin of 405 degrees is also root two over two. Multiplying 315 by three gives us 945. So, we need to work out the value of sin of 945 degrees. Two full turns is equal to 720 degrees. And adding 225 degrees to this, we see that 945 degrees lies in the third quadrant. The sin of this angle is therefore equal to negative root two over two as the sine of any angle in the third quadrant is negative.

We donβt need to consider three multiplied by negative 315 degrees or three multiplied by 405 degrees as these angles lie in the same standard position as 45 degrees. And the coefficient of the argument is an integer. We will therefore begin by substituting π₯ equals 45 degrees into our equation. The left-hand side becomes sin squared of 45 degrees plus six multiplied by sin of 135 degrees minus 0.5. From the values we calculated from our CAST diagram, this is equal to root two over two squared plus six multiplied by root two over two minus one-half. Root two over two squared is equal to two over four, which simplifies to one-half. So, we have one-half plus three root two minus one-half, which is equal to three root two.

We can therefore conclude that when π₯ is equal to 45 degrees, 405 degrees, or negative 315 degrees, then sin squared π₯ plus six sin three π₯ minus 0.5 is equal to three root two.

When π₯ is equal to 135 degrees, we have sin squared 135 degrees plus six multiplied by sin of 405 degrees minus 0.5. This is also equal to root two over two squared plus six multiplied by root two over two minus one-half, which once again simplifies to three root two. We can therefore also rule out 135 degrees. Finally, we consider π₯ is equal to 305 degrees. The left-hand side of the equation becomes sin squared of 305 degrees plus six multiplied by sin of 945 degrees minus 0.5. Using our values of sin of 315 and 945 degrees, we have negative root two over two squared plus six multiplied by negative root two over two minus one-half.

Squaring negative root two over two gives us two over four or one-half. And multiplying six by negative root two over two gives us negative three root two. Our expression simplifies to a half minus three root two minus a half. And this is equal to negative three root two. We can therefore conclude that the value of π₯ that does not satisfy the equation is option (D), 315 degrees.

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