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1 year ago

Help! How would I solve this trig identity?

Mathematics

1 answer:

NeTakaya1 year ago

7 0

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

$sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}$

Then the identity can be rewritten as:

$sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\$

Now we can multiply both sides by cos⁴(x) to get:

$\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)$

Now we can use the identity:

sin²(x) + cos²(x) = 1

$1 - cos^2(x) = sin^2(x)*(sin^2(x) + cos^2(x)) = sin^2(x)\\\\1 = sin^2(x) + cos^2(x) = 1$

Thus, the identity was proven.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

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3 years ago

A report by the US Geological Survey indicates that glaciers in Glacier National Park, Montana, are shrinking. Recent estimates

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Answer

a) C

b) B

c) C, E

d) 15.5

e) 0.7

f) 67.7

Step-by-step explanation

a) A linear equation can be written as $y=mx+c$ where $m$ is the slope and $c$ is the intercept on the $y$ -axis. The slope describes how $y$ changes when $x$ changes. A negative value means the it is a decreasing change. In our case, $y$ is $A$ and $x$ is $t$ . Thus, the slope of the equation is $m=-0.062$ . This means that the area of glacier is decreasing by $0.062 \text{km}{}^2 = 62000 \text{m}{}^2$ every year since 2000.

b) The $A$ -intercept (= 16.2 $\text{km}{}^2$ ) represents the area covered when $t=0$ . But $t=0$ starting from year 2000. This intercept then represents the area covered by glacier in the year 2000. Therefore, the area covered by glacier in 2000 was 16.2 $\text{km}{}^2$ .

c) $A=f(t)$ means the area covered after year 2000. Setting it to 12 means the area covered after $t$ years is 12 $\text{km}{}^2$ ). $t$ is therefore the number of years since 2000 that the total area covered will be 12 $\text{km}{}^2$ ).

d) $f(12)=16.2−0.062\times12 =16.2-0.744 = =15.456 15.5 \text{km}{}^2$ to 1 decimal place.

e) The term involving $t$ represents the disappearing area. In 12 years, the disappeared area is $0.062\times12=0.744 =0.7 \text{km}{}^2$ to 1 decimal place.