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dimaraw [331]
4 years ago
13

If we consider the toss of four coins as an experiment, how many equally likely outcomes does the sample space have?

Mathematics
1 answer:
Arturiano [62]4 years ago
6 0
There are 16 possible outcomes.

This is found by taking the number of possible outcomes for each toss, 2, and raising it to the 4th power (for 4 tosses):

2⁴ = 2*2*2*2 = 16
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<h3><u>Answer</u> :</h3>

We know that,

\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}

\dag\bf\:sin2A=2sinA\:cosA

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

<u>Now, Let's solve</u> !

\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}

\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}

\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}

\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}

\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}

\leadsto\:\bf{tan(A+B)}

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