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3 years ago

Verify sin(360 - etheta = -sin etheta

Mathematics

1 answer:

Dovator [93]3 years ago

8 0

Answer:

see explanation

Step-by-step explanation:

Using the subtraction identity for sine

sin(a + b) = sinacosb - cosasinb

Given

sin(360 - Θ)°

= sin360°cosΘ° - cos360°sinΘ°

= (0 × cosΘ ) - (1 × sinΘ)

= 0 - sinΘ

= - sinΘ ← as required

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Natasha2012 [34]

Answer:

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Step-by-step explanation:

We have been a quadratic equation:

$2x^2-10x-3$

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

$x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}$

General form of quadratic equation is $ax^2+bx+c$

On comparing general equation with b given equation we get

a=2,b=-10,c=-3

On substituting the values in formula we get

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Now substituting D in $x=\frac{b^2\pm\sqrt{D}}{2a}$ we get

$x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}$

$x=\frac{100\pm\sqrt{124}}{4}$

$x=\frac{100\pm2\sqrt{31}}{4}$

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Therefore, $x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}$

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