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3 years ago

Which expression is equivalent to sin(1.8x) sin(0.5x)?

1 answer:

3 0

Hum, this problem was difficult. You use the next expression to solve this problem. \[\cos (A - B) = \cos A \cos B + \sin A \sin B \] \[\cos (A + B) = \cos A \cos B - \sin A \sin B\] \[\cos (A - B ) - \cos (A +B ) =2 \sin A \sin B\] So \[\sin A \sin B = 0.5 \left( \cos(A - B) - \cos(A + B) \right)\] A = 1.8 x, B = 0.5 x \[\sin (1.8x) \sin (0.5x) = 0.5\left( \cos(1.8-0.5)x - \cos(1.8+0.5)x \right)\]\[= 0.5 \left( \cos(1.3x) - \cos (2.3x) \right)\] It's finish !!

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3 years ago

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2 years ago

Read 2 more answers

Find the x-intercepts for the parabola defined by the equation below. y = x2 + 9x + 14

faltersainse [42]

Answer:

x - intercepts are (-7,0) and (-2,0)

A is the correct option.

Step-by-step explanation:

We have been given the equation of the parabola $y=x^2+9x+14$

For x- intercept, y = 0

$0=x^2+9x+14$

We can split the middle term as 9x = 7x +2x

$x^2+7x+2x+14=0$

Now take GCF

$x(x+7)+2(x+7)=0$

Factored out the common term

$(x+7)(x+2)=0$

Apply the zero product property

$(x+7)=0,(x+2)=0$

Solve for x

$x=-7,-2$

Hence, x - intercepts are (-7,0) and (-2,0)

A is the correct option.

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B

inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$