Respuesta :
We want to know if the following statements are true or false
item a)
[tex]1-\cos x=\sin x[/tex]To solve this one, we're going to use the following identity
[tex]\sin ^2x+\cos ^2x=1[/tex]Rewriting this expression "isolating" the sine, we have
[tex]\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \sin ^2x=1-\cos ^2x \\ \sin x=\pm\sqrt[]{1-\cos^2x} \end{gathered}[/tex]Using this in our expression, we have
[tex]\begin{gathered} 1-\cos x=\sin x \\ \Rightarrow1-\cos x=\sqrt[]{1-\cos^2x} \end{gathered}[/tex]When we have a difference of two squares, we can separate them as
[tex]a^2-b^2=(a+b)(a-b)[/tex]Using this in the argument of the square root, we have
[tex]1-\cos ^2x=(1+\cos x)(1-\cos x)[/tex]Using this to rewrite our expression
[tex]\begin{gathered} 1-\cos x=\sqrt[]{1-\cos^2x} \\ 1-\cos x=\sqrt[]{(1+\cos x)(1-\cos x)} \\ 1-\cos x=\sqrt[]{(1+\cos x)}\sqrt[]{(1-\cos x)} \\ \frac{1-\cos x}{\sqrt[]{1-\cos x}}=\sqrt[]{1+\cos x} \end{gathered}[/tex]We can rewrite the numerator of the left side of the equation as the product of its square root.
[tex]1-\cos x=(\sqrt[]{1-\cos x})^2[/tex]Using this in our expression, we have
[tex]\begin{gathered} \frac{(\sqrt[]{1-\cos x})^2}{\sqrt[]{1-\cos x}}=\sqrt[]{1+\cos x} \\ \sqrt[]{1-\cos x}=\sqrt[]{1+\cos x} \\ 1-\cos x=1+\cos x \\ -\cos x=\cos x \end{gathered}[/tex]Since the cosine of an angle is not equal to minus its value, the first statemente is FALSE.
item b)
[tex]1-\cos ^2x=\sin ^2x[/tex]We're going to use the same identity to solve this one.
[tex]\sin ^2x+\cos ^2x=1[/tex]If we substitute the number 1 in our statement for this identity, we're going to have
[tex]\begin{gathered} 1-\cos ^2x=\sin ^2x \\ (\sin ^2x+\cos ^2x)-\cos ^2x=\sin ^2x \\ \sin ^2x+\cos ^2x-\cos ^2x=\sin ^2x \\ \sin ^2x=\sin ^2x \\ 1=1 \end{gathered}[/tex]Since we got a true equation in the end, the statement on this item is TRUE.
item c)
[tex]\frac{1}{\sin x}=\csc x[/tex]This is a given identity(you can check it is a part of the table above the question), then, it is TRUE.
item d)
[tex]\frac{\cos x}{\sin x}=\cos x(\frac{1}{\sin x})[/tex]To solve this one, we're going to use the following property.
[tex]\frac{a}{b}=a\cdot\frac{1}{b}[/tex]You can take out the numerator as a coefficient for our fraction.
Then, this statement is also TRUE.
item e)
[tex]\frac{\cos x}{\sin x}=\cos x\csc x[/tex]Using the previous two statements, we can solve this one,
Since the statement d is true
[tex]\frac{\cos x}{\sin x}=\cos x(\frac{1}{\sin x})[/tex]And statement c is also true
[tex]\frac{1}{\sin x}=\csc x[/tex]Then, if we substitute c on d, we have statement e, therefore, statement e is also TRUE.
item f)
[tex]\cos ^2x-1=\sin ^2x[/tex]
To solve this one, we're going to use the following identity
[tex]\cos ^2x+\sin ^2x=1[/tex]Again, making a substitution, we have
[tex]\begin{gathered} \cos ^2x-1=\sin ^2x \\ \cos ^2x-(\sin ^2x+\cos ^2x)=\sin ^2x \\ \cos ^2x-\sin ^2x-\cos ^2x=\sin ^2x \\ -\sin ^2x=\sin ^2x \end{gathered}[/tex]Since the last equation is false, then the statement f is also FALSE.
item g)
[tex]\frac{2}{\cos x}=2\sec x[/tex]Using the same property used on item d, we have
[tex]\frac{2}{\cos x}=2(\frac{1}{\cos x})[/tex]And also using the definition os sec x, we have
[tex]\sec x=\frac{1}{\cos x}[/tex]Using those properties in our statement, we have
[tex]\begin{gathered} \frac{2}{\cos x}=2\sec x \\ 2(\frac{1}{\cos x})=2\sec x \\ 2(\sec x)=2\sec x \\ 2\sec x=2\sec x \end{gathered}[/tex]Then, our last statement g is also TRUE.
