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Novay_Z [31]

Answer:

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

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

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

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

(C) Tan(A + B) = $\frac{Sin(A +B)}{Cos(A + B)}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

3 0

3 years ago

PLS HELP, THANK YOUU

ruslelena [56]

Jesos it’s all pixelated I can’t see anything on mi phone br

7 0

3 years ago

Read 2 more answers

if in a drug-study to test the effectiveness of a new drug for treating the common cold, you are randomly assigned to a group th

Free_Kalibri [48]

This means that you are in the experiment group

It is common to choose a group of people to study the effect of certain things like the one given in this question. The selection of people is done after designing the sample size for a certain population. The design varies depending on whether the analysis is qualitative or quantitative.

The people who lie in the sample size actually contribute to the experiment are said to be part of it.

Therefore, if you are part of the study that is used to test the effectiveness of a new drug for treating the common cold. It means you are part of the experiment group.

You can learn more about the experiment groups and control groups from

brainly.com/question/24708013
#SPJ4

3 0

2 years ago

Suppose you have two different linear equations. How would you tell if the lines represented are parallel, perpendicular, or nei

RoseWind [281]

Parallel never cross
perpendicular is a 90 degree angle
its neither if its not 90 degree or never crossing

8 0

3 years ago

2. 2x+6y=6<br> -x-3y = -3<br> Solution(s):<br> PLEASE HELP SOLVE FOR X AND Y

allochka39001 [22]

Answer:

x = 0, y = 1

Step-by-step explanation:

2x + 6y = 6 → (1)

x - 3y = - 3 ( add 3y to both sides )

x = 3y - 3 → (2)

substitute x = 3y - 3 into (1)

2(3y - 3) + 6y = 6 ← distribute parenthesis and simplify left side

6y - 6 + 6y = 6

12y - 6 = 6 ( add 6 to both sides )

12y = 12 ( divide both sides by 12 )

y = 1

substitute y = 1 into (2)

x = 3(1) - 3 = 3 - 3 = 0

solution is x = 0 , y = 1

3 0

2 years ago