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solong [7]
3 years ago
13

Simplify: (sin Θ − cos Θ)^2 + (sin Θ + cos Θ)^2

Mathematics
2 answers:
Ede4ka [16]3 years ago
5 0
Hope this could help you

Sidana [21]3 years ago
3 0

Answer:  The required simplified value of the given expression is 2.

Step-by-step explanation:  We are given to simplify the following trigonometric expression :

T_E=(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2~~~~~~~~~~~~~~~~~~~~~~~~~(i)

We will be using the following formulas :

(i)~(a-b)^2=a^2-2ab+b^2,\\\\(ii)~(a+b)^2=a^2+2ab+b^2,\\\\(iii)~\sin^2\theta+\cos^2\theta=1.

From (i), we have

T_E\\\\=(\sin\theta-\cos\theta)^2+(\sin\theta+\cos\theta)^2\\\\=\sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta+\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta\\\\=2(\sin^2\theta+\cos^2\theta)\\\\=2\times1\\\\=2.

Thus, the required simplified value of the given expression is 2.

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(\frac{1}{5}x+1)(\frac{1}{6}x+\frac{5}{4})

Use distributive property to remove the parentheses:

(a+b)(c+d)=ac+ad+bc+bd\begin{gathered} =\frac{1}{5}x*\frac{1}{6}x+\frac{1}{5}x*\frac{5}{4}+1*\frac{1}{6}x+1*\frac{5}{4} \\  \\ Multiplication\text{ }of\text{ }fractions: \\ \frac{a}{b}*\frac{c}{d}=\frac{a*c}{b*d} \\  \\ =\frac{1}{30}x^2+\frac{5}{20}x+\frac{1}{6}x+\frac{5}{4} \\  \\ =\frac{1}{30}x^2+\frac{1}{4}x+\frac{1}{6}x+\frac{5}{4} \\  \\ Addition\text{ }of\text{ }fractions: \\ \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd} \\  \\ =\frac{1}{30}x^2+(\frac{6x+4x}{24})+\frac{5}{4} \\  \\ =\frac{1}{30}x^2+\frac{10}{24}x+\frac{5}{4} \\  \\ Simplify: \\ =\frac{1}{30}x^2+\frac{5}{12}x+\frac{5}{4} \end{gathered}Then, the product in the simplest form is:\frac{1}{30}x^2+\frac{5}{12}x+\frac{5}{4}

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Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis
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Answer:

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

f(x)=6

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

f(x)=-3

The value of y is 1 at x=2, because of he close circles on (2,1).

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f(x)=1

Therefore the graph represents a piecewise function, which is defined as

f(x)=\begin{cases}6& \text{ if } x2 \end{cases}

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

lim_{x\rightarrow 2^-}f(x)=6

lim_{x\rightarrow 2^+}f(x)=-3

6\neq-3

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.

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Answer:

Step-by-step explanation:

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