sin^2 a(sq root(cos a))-sq root(cos^7 a)o) Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that k>0 and 0let x=ksin theta in sq root(k^2-x^2). Then find cos theta and tan theta. 1. cos theta=2. tan theta= p) Verify the identity.......
Choose the correct simplified form of sq root(cos a sin^4 a)-sq root(cos^7 a).
1. (sin^2 a-cos^3 a)sq root(cos a)
2. sq root(cos a sin^4 a-cos^7 a)
3. sq root(cos a sin^4 a-cos^7 a)-cos^3 a
4. sin^2 a(sq root(cos a))-sq root(cos^7 a)
o) Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that k>0 and 0
let x=ksin theta in sq root(k^2-x^2). Then find cos theta and tan theta.
1. cos theta=
2. tan theta=
p) Verify the identity. (csc t-1/cot t)=(cot t/csc t+1).
Show the sequence of steps that verifies the identity.
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If you could help in any way, that would be wonderful!!!
I'll do a couple of them for you.
i) Determine if the equation is an identity (cot(x) - tan(x)) / (sec(x)csc(x))=1. Yes or No?
An equation is an identity if it holds true for all values of x (and y and z, and whatever other variables are in there) where both sides of the equation are defined. If something is not an identity, it's usually pretty easy to find a value of x for which it does not hold true, so let's start by assuming it's not, and finding proof. The way you prove something is not an identity is you find a value of x for which the left hand side does not equal the right hand side.
Normally I'd try x = 0 or x = pi/2 first, because the trig values are so easy to compute, but unfortunately, tan(x) is not defined for x = pi/2, and cot(x) is not defined for x = 0, so we'll have to move on to my second choice, x = pi/4. When x = pi/4, we have:
RHS = 1 ... (as always!)
LHS = (cot(pi/4) - tan(pi/4)) / (sec(pi/4)csc(pi/4))
= (1 - 1) / (1/sqrt(2) * 1/sqrt(2))
= 0
Aha! So, RHS does not equal LHS for this value of x! The equation is not an identity.
f) Factor and simplify.
tan^3(x) + 8 = ?
To do this, use the sum of two cubes factorisation:
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
In this case a = tan(x), b = 2. We get:
tan^3(x) + 2^3 = (tan(x) + 2)(tan^2(x) - 2tan(x) + 4)
Hope that helps!
