super hard trig question i need answered, please!?

Question

An angles that measures 7x devided by 4 radians is in standard position. In which quadrant does its terminal side lie?

Answer 1

Depends on what x equals. The quadrants are bounded by multiples of pi/2. the first quadrant goes from 0 to pi/2, the second goes from pi/2 to pi, the third from pi to 3pi/2, and the last from 3pi/2 to 2pi.

If your angle is 7x/4, you'd need to know what x is to determine where this angle lies. If it's greater than 2pi, then start adding pi/2 to everything for new bounds for your angle - 2pi to 5pi/2 for first quadrant, 5pi/2 to 3 pi for second, etc.

If by chance x = pi, then your terminal side would be on the negative y axis - you would have a straight angle, and your terminal side wouldn't be in a quadrant, since it's on the boundary between two quadrants.

Answer 2

sin²? + cos²? + tan²? + sec²? + csc²? + cot²? = 7 (sin²? + cos²?) + tan²? + (a million + tan²?) + (a million + cot²?) + cot²? = 7 a million + tan²? + (a million + tan²?) + (a million + cot²?) + cot²? = 7 3 + 2tan²? + 2cot²? = 7 2tan²? + 2cot²? = 4 tan²? + cot²? = 2 sin²?/cos²? + cos²?/sin²? = 2 (sin?? + cos??) / sin²?cos²? = 2 sin?? + cos?? = 2sin²?cos²? sin?? + -2sin²?cos²? + cos?? = 0 (sin²? - cos²?)² = 0 sin²? - cos²? = 0 sin²? - (a million - sin²?) = 0 2sin²? - a million = 0 sin²? = a million/2 sin? = ±?2/2 ? = ?/4 + ?k/2, the place ok is an integer. for ? [0, 2?): ?/4, 3?/4, 5?/4, 7?/4 in radians. for [0°? ? < 360°): {?: 40 5°, one hundred thirty five°, 225°, 315°} i desire this facilitates!