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2 years ago

Prove sin^2 theta-sin^2 theta is sin tetha sin 3 thetahow to prove..pls help!! thanksss

Mathematics

1 answer:

Artemon [7]2 years ago

3 0

Simply you can substitute theta=45 degrees (or any angle,not zero) in left hand side and right hand side.

or

using a^2 - b^2 =(a+b)(a-b)

= sin(2x+x)sin(2x-x)
= sin 3x sin x

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Write an explicit formula for the An, and the n^th term of the sequence 24,-12,6,....

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Answer:

$T_n = -48* (-\frac{1}{2})^{n}$

Step-by-step explanation:

Given

$24, -12, 6,...$

Required

Write a formula

The above sequence shows a geometric sequence because:

$r = \frac{-12}{24} = \frac{6}{-12} = -\frac{1}{2}$ -- common ratio

The equation is determined using:

$T_n = ar^{n-1}$

Where

$a = 24$

Substitute values for a and r

$T_n = 24* (-\frac{1}{2})^{n-1}$

Apply law of indices:

$T_n = 24* (-\frac{1}{2})^{n} * (-\frac{1}{2})^{-1}$

$T_n = 24* (-\frac{1}{2})^{n} * 1/(-\frac{1}{2})$

$T_n = 24* (-\frac{1}{2})^{n} * 1*-2$

$T_n = 1*-2*24* (-\frac{1}{2})^{n}$

$T_n = -48* (-\frac{1}{2})^{n}$

<em>The above represents the explicit formula</em>

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Answer:

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