Find the binomial expansion of √1x in increasing power of x up to the term in x⁴
(b)(i) prove that the trigonometric identity_____1_____ _____cot²x______=cos²(1 tan²x) ² (1 cot²x) ²
(b)(ii) Find all values of Z=64³?
If you mean √(1+x) then that's
(1+x)^(1/2) = 1^(1/2)*x^0 + (1/2)*1^(-1/2)*x^1 + ... + (1/2)(-1/2)(-3/2)(-5/2)*1^(-7/2)*x^4 + ...
= 1 + x/2 - x^2/8 + x^3/16 - 5x^4/128 + ...
(bi) How about some + and - signs in there? If there's a fraction involved, use / and () to disambiguate.
(bii) I assume you mean 64^(1/3).
That would be 4cis(2kπ/3) for all integer k
How about a complete expression of the problem solution
To find the binomial expansion of √(1 + x), we can use the Binomial Theorem. The Binomial Theorem states that for any real number "n" and any real number "r", the expansion of (a + b)^n can be found using the formula:
(a + b)^n = C(n,0) a^n b^0 + C(n,1) a^(n-1) b^1 + C(n,2) a^(n-2) b^2 + ... + C(n,r) a^(n-r) b^r + ... + C(n,n) a^0 b^n
Where C(n, r) is the binomial coefficient given by:
C(n, r) = n! / (r!(n-r)!)
In our case, a = 1 and b = x, and we want to find the expansion of √(1 + x) up to the term in x^4. So, we'll substitute these values into the formula and determine the respective coefficients.
√(1 + x) = C(1/2, 0) 1^(1/2) x^0 + C(1/2, 1) 1^(1/2-1) x^1 + C(1/2, 2) 1^(1/2-2) x^2 + C(1/2, 3) 1^(1/2-3) x^3 + C(1/2, 4) 1^(1/2-4) x^4
Simplifying using the binomial coefficients and simplifying the other terms:
√(1 + x) = 1 + (1/2)x - (1/8)x^2 + (1/16)x^3 - (5/128)x^4
Therefore, the binomial expansion of √(1 + x) up to the term in x^4 is 1 + (1/2)x - (1/8)x^2 + (1/16)x^3 - (5/128)x^4.
Now, moving on to part (i) of question (b):
To prove the trigonometric identity (1 + cot²x)² = cos²(1 + tan²x), we can start by simplifying both sides of the equation individually.
Left-hand side (LHS): (1 + cot²x)²
Expanding and simplifying:
LHS = (1 + cot²x)(1 + cot²x)
= 1 + 2cot²x + cot⁴x
Right-hand side (RHS): cos²(1 + tan²x)
Expanding and simplifying:
RHS = cos²(1 + tan²x)
= cos²(1 + sin²x/cos²x)
= cos²(cos²x + sin²x)/cos²x
= cos⁴x/cos²x
= (cos²x)²/cos²x
= cos²x
As we can see, the LHS and RHS of the equation are equal. Hence, the trigonometric identity (1 + cot²x)² = cos²(1 + tan²x) is proven.
Moving on to part (ii) of question (b):
To find the value of Z = 64³, we simply calculate the cube of 64.
Z = 64³
= 64 * 64 * 64
= 262,144
Therefore, the value of Z = 64³ is 262,144.