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2 years ago

(sin30° + cos30°) – (sin 60° + cos60°)

Mathematics

1 answer:

Darya [45]2 years ago

8 0

Answer:

Step-by-step explanation:

Sin 30° = ½

Cos 30° = √3/2

Sin 60° = √3/2

Cos 60° = ½

Therefore,

(½ + √3/2) - (√3/2 + ½)

$\frac{1 + \sqrt[]{3} }{2} - \frac{ \sqrt{3 + 1} }{ 2 }$

$\frac{1 + \sqrt{3} - \sqrt{3} - 1}{2}$

$\frac{0 - 0}{2}$

Answer: 0

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What does y equal in this problem 5y-3(2-y)=10

Pepsi [2]

Steps:

5y - 6 + 3y = 10

8y = 10 + 6

y = 16/8

y = 2

Hope this helps! :)

3 0

3 years ago

3(2x-7=5-(1-x) pls help me

vekshin1

Answer:

3 • (2x - 7) - (x + 4) = 0

then do 3.1 Pull out like factors :

5x - 25 = 5 • (x - 5)

5 • (x - 5) = 0

4.2 Solve : x-5 = 0

Add 5 to both sides of the equation :

x = 5

Step-by-step explanation:

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2 years ago

In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc

Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

$A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°$

Use sine rule,

$\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}$

Again consider,

$\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Thus, the angle B is function of A is, $B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

Now find $\frac{dB}{dA}$

Differentiate implicitly the function $\sin{B}=\sqrt{\frac{3}{2}}\sin{A}$ with respect to A to get,

$\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}$

When $A=\frac{\pi}{4},B=\frac{\pi}{3}$ , the value of $\frac{dB}{dA}$ is,

$\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}$

In general, the linear approximation at x= a is,

$f(x)=f'(x).(x-a)+f(a)$

Here the function $f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]$

At $A=\frac{\pi}{4}$

$f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}$