Simplify the trigonometric expression sec theta cos theta a. tan theta b. 1 c. cot theta d. sin theta
1 answer:
Question:<u /> <em>Simplify the trigonometric expression sec</em><span><em>Өcos</em></span><span><em>Ө.</em></span>
For each of the three basic trig functions, (sin, cos, tan) there are three more which act as their reciprocals. (csc, sec, and cot respectively)
Reciprocals, of course, are flipped fractions. If you have some number x, the reciprocal can be expressed as
.
An important property of reciprocals is that if you mutliply them together they will cancel out--take
for example.
Since sec<span>Ө is literally <em>defined</em> as the reciprocal of cos</span><span>Ө, it is evident that
</span>
For each of the three basic trig functions, (sin, cos, tan) there are three more which act as their reciprocals. (csc, sec, and cot respectively)
Reciprocals, of course, are flipped fractions. If you have some number x, the reciprocal can be expressed as
An important property of reciprocals is that if you mutliply them together they will cancel out--take
Since sec<span>Ө is literally <em>defined</em> as the reciprocal of cos</span><span>Ө, it is evident that
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Pictures always help; see the one I've attached for reference.
Call the first vector (from HQ to the supply drop) u and the second vector (from supply drop to medics) v. We then want to find w, the vector from the medics to HQ, which corresponds to the vector -(u + v). (This is because u + v is the vector pointing from HQ to the medics; we're talking about the one in the opposite direction.)
Write the vectors in horizontal/vertical component form:
u = (125 km) (cos 235º x + sin 235º y) = (-71.70 x - 102.39 y) km
v = (75 km) (cos 110º x + sin 110º y) = (-25.65 x + 70.48 y) km
Why these angles?
- "55 degrees south of west" is 235º; due west is 180º from the positive horizontal axis, and you add 55º to this
- "20 degrees west of north" is 110º; due north is 90º, so add 20º to this
Add the vectors:
u + v = (-97.35 x - 31.92 y) km
Multiply by -1 to get the vector w:
w = -(u + v) = (97.35 x + 31.92 y) km
The distance covered by this vector is equal to its magnitude:
||w|| = √((97.35 km)^2 + (31.92 km)^2) = 102.45 km
The direction <em>θ</em> is given by
tan<em>θ</em> = (31.92 km)/(97.35 km) ==> <em>θ</em> = 18.15º
