All categories

2 years ago

Write expression as a sine or cosine of an angle.cos 94° cos 18° + sin 94° sin 18°

Mathematics

1 answer:

MArishka [77]2 years ago

6 0

<h3>Answer: cos(76)</h3>

=========================================================

Explanation:

The original expression is of the pattern cos cos + sin sin. This pattern matches the second identity in the hint. Specifically, we'll say the following:

cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

cos(A)cos(B) + sin(A)sin(B) = cos(A - B)

cos(94)cos(18) + sin(94)sin(18) = cos(94 - 18)

cos(94)cos(18) + sin(94)sin(18) = cos(76)

----------

We can verify this by use of a calculator. Make sure your calculator is in degree mode.

cos(94)cos(18) + sin(94)sin(18) = 0.24192
cos(76) = 0.24192

Both expressions give the same decimal approximation, so this helps confirm the two expressions are equal. You could also use the idea that if x = y, then x-y = 0. Through this method, you'll subtract the left and right hand sides and you should get (very close to) zero.

You might be interested in

Use f(x)=1/2x and f^-1(x)=2x to solve the problems f^1(-2), f(-4), f(f^-1(-2)?

neonofarm [45]

F^1(-2) = 2(-2) = -4
f(-4) = 1/2 * -4 = -2

f(f^-1(x)) = x so answer is -2

8 0

2 years ago

Read 2 more answers

Need help plz very much

andrew-mc [135]

Answer:

Step-by-step explanation:

Remember that when we are dividing two exponents with the same base (here, it's 13), we can subtract the values of the powers:

$\frac{a^m}{a^n} =a^{m-n}$

So, here, we have $\frac{13^{10}}{13^7}$ , so we subtract 7 from 10:

$\frac{13^{10}}{13^7}=13^{10-7}=13^3$

Thus, 3 should go in the box.

6 0

2 years ago

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi

lara [203]

For each curve, plug in the given point $(x,y)$ and check if the equality holds. For example:

(I) (2, 3) does lie on $x^2+xy-y^2=1$ since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative $\frac{\mathrm dy}{\mathrm dx}$ , and evaluate it for the given point $(x,y)$ . This is the slope of the tangent line at the point. For example:

(I) The derivative is

$x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}$

so the slope of the tangent at (2, 3) is

$\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74$

and its equation is then

$y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12$

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, $-\frac1{\frac{\mathrm dy}{\mathrm dx}}$ . For example: