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deff fn [24]
3 years ago
14

IF A+B+C=Π prove that sin(3A)+sin(3B)+sin(3C)=-4cos(3A÷2)cos(3B÷2)cos(3C÷2)​

Mathematics
1 answer:
Hunter-Best [27]3 years ago
4 0

Answer:

LHS of the given equation is:

sinθ+sin3θ+sin5θ+sin7θ=(sinθ+sin7θ)+(sin3θ+sin5θ)=2sin8θ2.cos6θ2+2sin8θ2.cos2θ2   [since sinC+sinD=2sinC+D2.cosC−D2=2sin4θ.cos3θ+2sin4θ.cosθ=2sin4θ.[cos3θ+cosθ]       [since cosC+cosD=2cosC+D2.cosC−D2=2sin4θ.[2cos2θ.cosθ]=4cosθ.cos2θ.sin4θsinθ+sin3θ+sin5θ+sin7θ=(sinθ+sin7θ)+(sin3θ+sin5θ)=2sin8θ2.cos6θ2+2sin8θ2.cos2θ2   [since sinC+sinD=2sinC+D2.cosC-D2=2sin4θ.cos3θ+2sin4θ.cosθ=2sin4θ.[cos3θ+cosθ]       [since cosC+cosD=2cosC+D2.cosC-D2=2sin4θ.[2cos2θ.cosθ]=4cosθ.cos2θ.sin4θ

any more help just ask :)

Step-by-step explanation:

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3 years ago
The table shows information about the numbers of hours 30 children spent on their tablets one evening. a) find the class interva
Bogdan [553]

Answer:

a) 20<h≤30.

b) 26.17 hrs

Step-by-step explanation:

The missing table is shown in attachment.

Part a)

We need to find the class interval that contains the median.

The total frequency is

\sum \: f = 30

The median class corresponds to half

\frac{1}{2}  \sum \: f ^{th}  -  -  - value

That is the 15th value.

We start adding the frequency from the top obtain the least cumulative frequency greater or equal to 15.

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This corresponds to the class interval 20<h≤30.

Adding from the bottom also gives the same result.

Therefore the median class is 20<h≤30.

b) Since this is a grouped data we use the midpoint to represent the class.

The median is given by :

\frac{\sum \: fx}{ \sum\: f}

=  \frac{5 \times 3 + 15\times 8 + 25 \times 9 + 35 \times 7 + 45 \times 4}{30}

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3 years ago
Read 2 more answers
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